NCERT Solutions For Class 11 Physics Chapter 11

Q.1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Ans.

\begin{array}{l}\text{Relation}\text{between}\text{Kelvin and Celsius scales is}\text{given as:}\\ {\text{T}}_{\text{C}}{\text{= T}}_{\text{K}}\text{– 273}\text{.15}\left(\text{i}\right)\\ \text{Relation}\text{between}\text{Celsius and Fahrenheit scales is}\text{given as:}\\ {\text{T}}_{\text{F}}\text{=}\frac{\text{9}}{\text{5}}{\text{T}}_{\text{C}}\text{+ 32}\left(\text{ii}\right)\\ \text{In}\text{case}\text{of neon,}\\ {\text{T}}_{\text{K}}\text{=}\text{24}\text{.57 K}\\ \text{From}\text{equation}\text{(i),}\text{we}\text{obtain:}\\ {\text{T}}_{\text{C}}\text{=}\text{24}\text{.57}\text{–}\text{273}\text{.15 =}\text{–248}\text{.58°C}\\ \text{From}\text{equation}\text{(ii),}\text{we}\text{obtain:}\\ {\text{T}}_{\text{F}}\text{=}\frac{\text{9}}{\text{5}}{\text{T}}_{\text{C}}\text{+}\text{32}\\ \therefore {\text{T}}_{\text{F}}\text{=}\frac{\text{9}}{\text{5}}\left(\text{–248}\text{.58}\right)\text{+32}\\ \therefore {\text{T}}_{\text{F}}\text{= 415}{\text{.44}}^{\text{o}}\text{F}\\ \text{In}\text{case}\text{of carbon dioxide:}\\ {\text{T}}_{\text{K}}\text{= 216}\text{.55 K}\\ \text{From}\text{equation}\text{(i),}\text{we}\text{obtain:}\\ \therefore {\text{T}}_{\text{C}}\text{= 216}\text{.55}\text{–}\text{273}\text{.15 = –56}\text{.60°C}\\ \text{From}\text{equation}\text{(ii),}\text{we}\text{obtain:}\\ {\text{T}}_{\text{F}}\text{=}\frac{\text{9}}{\text{5}}{\text{T}}_{\text{C}}\text{+ 32}\\ \therefore {\text{T}}_{\text{F}}\text{=}\frac{\text{9}}{\text{5}}\left(\text{–56}\text{.60}\right)\text{+32 = -69}{\text{.88}}^{\text{o}}\text{F}\end{array}

Q.2 Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B?

Ans.

\begin{array}{l}{\text{Here, triple point of water on scale A, T}}_{\text{A}}\text{=}\text{200 A}\\ \text{Triple point of water on}{\text{scale B,T}}_{\text{B}}\text{= 350 B}\\ \text{Triple point of water on the}{\text{Kelvin scale, T}}_{\text{K}}\text{= 273}\text{.15 K}\\ \text{According}\text{to}\text{the}\text{question:}\\ \text{Temperature 273}\text{.15 K on Kelvin scale}\text{=}\text{200 A on}\text{absolute scale A}\\ {\text{T}}_{\text{1}}{\text{= T}}_{\text{K}}\\ \therefore \text{200}\text{A = 273}\text{.15 K}\\ \therefore \text{A =}\frac{\text{273}\text{.15 K}}{\text{200}}\\ \text{According}\text{to}\text{the}\text{question,}\\ \text{Temperature 273}\text{.15 K on Kelvin scale = 350 B on}\text{absolute scale B}\\ {\text{T}}_{\text{B}}{\text{= T}}_{\text{K}}\\ \therefore \text{350 B}\text{=}\text{273}\text{.15}\\ \therefore \text{B =}\frac{\text{273}\text{.15}}{\text{350}}\\ \text{Relation}\text{between}{\text{T}}_{\text{A}}\text{and}{\text{T}}_{\text{B}}\text{is}\text{given}\text{as:}\\ \therefore \frac{\text{273}\text{.15}}{\text{200}}{\text{× T}}_{\text{A}}\text{=}\frac{\text{273}\text{.15}}{\text{350}}{\text{× T}}_{\text{B}}\\ \therefore {\text{T}}_{\text{A}}\text{=}\frac{\text{200}}{\text{350}}{\text{T}}_{\text{B}}\\ \therefore \frac{{\text{T}}_{\text{A}}}{{\text{T}}_{\text{B}}}\text{=}\frac{\text{200}}{\text{350}}\\ \therefore \text{The ratio}{\text{T}}_{\text{A}}{\text{: T}}_{\text{B}}\text{= 4 : 7}\end{array}

Q.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R_o [1 + α (T – T_o)]. The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

Ans.

$\begin{array}{l}\text{Given}{\text{that, R = R}}_{\text{0}}{\text{[1 + α(T – T}}_{\text{0}}\text{)] (i)}\hfill \\ {\text{Here, R}}_{\text{0}}\text{and}{\text{T}}_{\text{0}}\text{represent}\text{initial resistance}\text{and}\text{initial}\text{temperature respectively}\text{.}\hfill \\ \text{R and}\text{T}\text{represent}\text{final resistance}\text{and}\text{final}\text{temperature respectively}\text{.}\hfill \\ \text{α = a constant}\hfill \\ \text{At triple point of water,}{\text{T}}_{\text{0}}\text{= 273}\text{.15 K}\hfill \\ \text{Resistance of the}\text{lead,}{\text{R}}_{\text{0}}\text{= 101}\text{.6 Ω}\hfill \\ \text{At the}\text{normal melting point of lead,}\text{T = 600}\text{.5 K}\hfill \\ \text{Resistance of the}\text{lead, R = 165}\text{.5 Ω}\hfill \\ \text{Putting these values in equation (i), we obtain:}\hfill \\ {\text{R=R}}_{\text{0}}{\text{[1 + α(T–T}}_{\text{0}}\text{)]}\hfill \\ \text{165}\text{.5 Ω = 101}\text{.6 Ω[1 + α(600}\text{.5 – 273}\text{.15)K]}\hfill \\ \text{1}\text{.629 Ω = 1 Ω + α(327}\text{.35)K}\hfill \\ \therefore \text{α =}\frac{\text{0}\text{.629}}{\text{327}\text{.35K}}\hfill \\ \text{= 1}{\text{.92 × 10}}^{\text{-3}}{\text{K}}^{\text{-1}}\hfill \\ \text{In}\text{case}\text{of resistance,}{\text{R}}_{\text{1}}\text{=123}\text{.4 Ω}\hfill \\ {\text{R}}_{\text{1}}{\text{= R}}_{\text{0}}\text{[1}{\text{+ α(T–T}}_{\text{0}}\text{)]}\hfill \\ \text{Here,}\text{T}\text{represents}\text{the}\text{temperature}\text{when}\text{resistance}\text{of}\text{lead}\text{=}\text{123}\text{.4 Ω}\hfill \\ \therefore \text{123}\text{.4 Ω = 101}\text{.6 [1 + 1}{\text{.92 ×10}}^{\text{-3}}{\text{K}}^{\text{-1}}\text{(T – 273}\text{.15)K]}\hfill \\ \text{1}\text{.214 Ω = 1Ω + 1}{\text{.92 ×10}}^{\text{-3}}{\text{K}}^{\text{-1}}\text{(T – 273}\text{.15)K}\hfill \\ \frac{\text{0}\text{.214 Ω}}{\text{1}{\text{.92×10}}^{\text{-3}}{\text{K}}^{\text{-1}}}\text{=}\text{T – 273}\text{.15}\hfill \\ \text{T = 384}\text{.61}\text{K}\hfill \end{array}$

Q.4  Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?
(c) The absolute temperature (Kelvin scale) T is related to the temperature t_c on the Celsius scale by
t_c = T – 273.15
Why do we have 273.15 in this relation, and not 273.16?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Ans.

\begin{array}{l}\left(\text{a}\right)\text{The triple point of water has a unique value}\text{. At the certain values}\\ \text{of volume and pressure, the triple point of water is always is equal to}\\ \text{273}\text{.16 K}\text{. The melting point of ice and the boiling point of water do not}\\ \text{have specific values because they depend on the pressure and temperature}\text{.}\\ \left(\text{b}\right)\text{The absolute zero or 0 K is the other stable point on the Kelvin scale}\text{.}\\ \left(\text{c}\right)\text{On Kelvin scale, the temperature 273}\text{.16 K is the triple point of water}\text{.}\\ \text{However, 273}\text{.16 K is not the melting point of ice}\text{. On Celsius scale, the}\\ \text{temperature 0°C is the melting point of ice and its equivalent value on Kelvin}\\ \text{scale is 273}\text{.15 K}\text{.}\\ \text{Therefore, on Celsius scale, the absolute temperature}\left(\text{Kelvin scale}\right)\text{T, is related}\\ {\text{to temperature t}}_{\text{c}}\text{, as:}\\ {\text{t}}_{\text{c}}\text{= T – 273}\text{.15}\\ {\text{d)Consider that T}}_{\text{K}}{\text{is the temperature on absolute scale and T}}_{\text{F}}\text{is the temperature on}\\ \text{Fahrenheit scale}{\text{. The temperatures T}}_{\text{F}}{\text{and T}}_{\text{K}}\text{can be related as:}\\ \frac{{\text{T}}_{\text{F}}\text{– 32}}{\text{180}}\text{=}\frac{{\text{T}}_{\text{K}}\text{– 273}\text{.15}}{\text{100}}\text{(i)}\\ {\text{Consider T}}_{\text{F1}}\text{be the temperature on Fahrenheit scale and}\\ {\text{T}}_{\text{K1}}\text{be the temperature on absolute scale}\text{.}\\ \text{The}{\text{temperatures T}}_{\text{F1}}\text{and}{\text{T}}_{\text{K1}}\text{can}\text{be related as:}\\ \frac{{\text{T}}_{\text{F1}}\text{– 32}}{\text{180}}\text{=}\frac{{\text{T}}_{\text{K1}}\text{– 273}\text{.15}}{\text{100}}\text{(ii)}\\ \text{Given that:}\\ {\text{T}}_{\text{K1}}{\text{–T}}_{\text{K}}\text{= 1 K}\\ \text{Subtracting}\text{equation}\text{(i)}\text{from}\text{equation}\text{(ii),}\text{we}\text{obtain:}\\ \frac{{\text{T}}_{\text{F1}}{\text{– T}}_{\text{F}}}{\text{180}}\text{=}\frac{{\text{T}}_{\text{K1}}{\text{– T}}_{\text{K}}}{\text{100}}\text{=}\frac{\text{1}}{\text{100}}\\ {\text{T}}_{\text{F1}}{\text{– T}}_{\text{F}}\text{=}\frac{\text{1 × 180}}{\text{100}}\text{=}\frac{\text{9}}{\text{5}}\\ \text{Triple}\text{point}\text{of}\text{water = 273}\text{.16}\text{K}\\ \therefore \text{Triple}\text{point}\text{of}\text{water}\text{on}\text{the}\text{new}\text{scale}\text{= 273}\text{.16 ×}\frac{\text{9}}{\text{5}}\text{= 491}\text{.69 K}\end{array}

Q.5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?

(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Ans.

\begin{array}{l}\text{Here,}\text{triple}\text{point}\text{of}\text{water,}\text{T = 273}\text{.16}\text{K}\\ \text{At}\text{the}\text{triple}\text{point}\text{of}\text{water,}\text{the}\text{pressure}\text{in}\text{thermometer}\text{A,}\\ {\text{P}}_{\text{A}}\text{= 1}{\text{.250 ×10}}^{\text{5}}\text{Pa}\\ \text{Let}\text{the}\text{normal}\text{melting}\text{point}\text{of}\text{sulphur}\text{=}{\text{T}}_{\text{1}}\\ \text{At}\text{the}\text{normal}\text{melting}\text{point}\text{of}\text{sulphur,}\text{the}\text{pressure}\text{in}\text{thermometer}\text{A,}\\ {\text{P}}_{\text{1}}\text{= 1}{\text{.797 × 10}}^{\text{5}}\text{Pa}\\ \text{As}\text{per}\text{Charles’}\text{law,}\text{we}\text{have:}\\ \frac{{\text{P}}_{\text{A}}}{\text{T}}\text{=}\frac{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{1}}}\\ \therefore {\text{T}}_{\text{1}}\text{=}\frac{{\text{P}}_{\text{1}}\text{T}}{{\text{P}}_{\text{A}}}\\ {\text{T}}_{\text{1}}\text{=}\frac{\text{1}{\text{.797×10}}^{\text{5}}\text{Pa × 273}\text{.16 K}}{\text{1}{\text{.250 ×10}}^{\text{5}}Pa}\text{= 392}\text{.69}\text{K}\\ \therefore \text{Absolute}\text{temperature}\text{of}\text{normal}\text{melting}\text{point}\text{of}\text{sulphur}\text{as}\\ \text{read}\text{by}\text{thermometer}\text{A}\text{=}\text{392}\text{.69}\text{K}\text{.}\\ \text{At}\text{triple}\text{point}\text{of}\text{water,}\text{pressure}\text{in}\text{thermometer}\text{B,}\\ {\text{P}}_{\text{B}}\text{=}\text{0}\text{.200}\text{×}{\text{10}}^{\text{5}}\text{Pa}\\ \text{At}\text{temperature}{\text{T}}_{\text{1}}\text{,}\text{pressure}\text{in}\text{thermometer}\text{B,}\\ {\text{P}}_{\text{2}}\text{=}\text{0}\text{.287}\text{×}{\text{10}}^{\text{5}}\text{Pa}\\ \text{As}\text{per}\text{Charles’}\text{law,}\text{we}\text{have,}\\ \frac{{\text{P}}_{\text{B}}}{\text{T}}\text{=}\frac{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{1}}}\\ \therefore {\text{T}}_{\text{1}}\text{=}\frac{{\text{P}}_{\text{2}}\text{T}}{{\text{P}}_{\text{B}}}\\ {\text{T}}_{\text{1}}\text{=}\frac{\text{0}{\text{.287×10}}^{\text{5}}\text{Pa × 273}\text{.16 K}}{\text{0}{\text{.200×10}}^{\text{5}}\text{Pa}}\\ {\text{T}}_{\text{1}}\text{= 391}\text{.98}\text{K}\\ \therefore \text{Absolute}\text{temperature}\text{of}\text{normal}\text{melting}\text{point}\text{of}\text{sulphur}\text{as}\\ \text{read}\text{by}\text{thermometer}\text{B}\text{=}\text{391}\text{.98}\text{K}\text{.}\\ \text{(b)}\text{The}\text{slight}\text{difference}\text{in}\text{the}\text{readings}\text{of}\text{thermometers}\text{A}\text{and}\text{B}\\ \text{is}\text{due}\text{to}\text{the}\text{fact}\text{that}\text{oxygen}\text{and}\text{hydrogen}\text{gases}\text{present}\text{in}\text{thermometers}\\ \text{A}\text{and}\text{B}\text{respectively}\text{are}\text{not}\text{perfectly}\text{ideal}\text{.}\text{To}\text{reduce}\text{the discrepancy}\\ \text{between}\text{the}\text{two}\text{readings,}\text{the}\text{readings}\text{should}\text{be}\text{taken}\text{at}\text{lower}\text{pressure}\\ \text{conditions}\text{as in}\text{that}\text{case,}\text{the}\text{gases}\text{approach}\text{to}\text{the}\text{ideal}\text{gas}\text{behaviour}\text{.}\end{array}

Q.6 A steel tape 1 m long is correctly calibrated for a temperature of 27.0 ^oC. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0^oC. What is the actual length of the steel rod on that day? What is the length of the same steel rod on the day when the temperature is 27.0 ^oC? Coefficient of linear expansion of steel = 1.20 × 10^{-5 o}C^-1.

Ans.

\begin{array}{l}\text{Here,}\text{length}\text{of}\text{steel}\text{tape}\text{at}\text{temperature}\text{T =}\text{27°C,}\text{l = 1}\text{m = 100}\text{cm}\\ \text{At}\text{temperature}{\text{T}}_{\text{1}}\text{=}{\text{45}}^{\text{o}}\text{C,}\text{length}\text{of}\text{the}\text{steel}\text{rod,}{\text{l}}_{\text{1}}\text{= 63}\text{cm}\\ \text{Coefficient}\text{of}\text{linear}\text{expansion}\text{of}\text{the}\text{steel,}\alpha \text{= 1}{\text{.20×10}}^{\text{-5}}{\text{K}}^{\text{-1}}\\ \text{Let}\text{the}\text{actual}\text{length}\text{of}\text{the}\text{steel}{\text{rod = l}}_{\text{2}}\\ \text{Let}\text{the}\text{length}\text{of}\text{the}\text{steel}\text{rod}\text{at}{\text{45}}^{\text{o}}\text{C = l’}\\ \text{l’ = l +}\alpha \text{l}\left({\text{T}}_{\text{1}}\text{– T}\right)\\ \therefore \text{l’ = 100 cm + 1}{\text{.20×10}}^{\text{-5}}{\text{K}}^{\text{-1}}\text{× 100 cm}\left(\text{45 – 27}\right)\text{°C}\\ \text{l’ = 100}\text{.0216}\text{cm}\\ \therefore \text{Actual}\text{length}\text{of}\text{the}\text{steel}\text{rod}\text{measured}\text{by}\text{the}\text{steel}\text{tape}\\ \text{at}\text{45°C,}{\text{l}}_{\text{2}}\text{=}\frac{\text{100}\text{.0216 cm}}{\text{100 cm}}\text{× 63 cm = 63}\text{.0136}\text{cm}\\ \therefore \text{Actual}\text{length}\text{of}\text{the}\text{steel}\text{rod}\text{at}\text{45°C}\text{is}\text{63}\text{.0136}\text{cm}\\ \text{and}\text{its}\text{length}\text{at}\text{27°C}\text{is}\text{63}\text{.0}\text{cm}\text{.}\end{array}

Q.7 A large steel wheel is to be lifted on a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range:

α_steel = 1.20×10^-5 K^-1.

Ans.

\begin{array}{l}\text{Here, temperature, T}\text{=}\text{27°C = 27°C + 273}\text{=}\text{300 K}\\ {\text{Outer diameter of the shaft at T, d}}_{\text{1}}\text{= 8}\text{.70 cm}\\ {\text{Diameter of the central hole in the wheel at T, d}}_{\text{2}}\text{=}\text{8}\text{.69}\text{cm}\\ \text{Coefficient of linear expansion of steel,}{\alpha }_{\text{s}}\text{=1}\text{.20}\text{×}{\text{10}}^{\text{–5}}{\text{K}}^{\text{–1}}\\ \text{After cooling,}\text{let}\text{temperature}\text{of}\text{the}\text{shaft}\text{=}{\text{T}}_{\text{1}}\\ \text{For the}\text{wheel to slip on the shaft,}\text{the change in diameter}\\ \text{of}\text{the}\text{shaft, Δd = 8}\text{.69 cm – 8}\text{.70 cm = –0}\text{.01 cm}\\ {\text{Temperature T}}_{\text{1}}\text{can be obtained from the relation:}\\ {\text{Δd = d}}_{\text{1}}{\alpha }_{\text{s}}\left({\text{T}}_{\text{1}}\text{–T}\right)\\ \text{0}\text{.01 cm = 8}\text{.70 cm × 1}{\text{.20 × 10}}^{\text{–5}}{\text{K}}^{-1}\left({\text{T}}_{\text{1}}\text{–300 K}\right)\\ \therefore {\text{T}}_{\text{1}}\text{–300K = 95}\text{.78 K}\\ \therefore {\text{T}}_{\text{1}}\text{= 204}\text{.21 K}\\ {\text{T}}_{\text{1}}\text{= 204}\text{.21 K – 273}\text{.16 K}\\ \text{= – 68}\text{.95 K}\\ \therefore \text{The wheel will slip on the shaft when the temperature}\\ \text{of the shaft is}\text{approximately}\text{–69 K}\text{.}\end{array}

Q.8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of copper = 1.70 × 10^–5 K^–1.

Ans.

\begin{array}{l}\text{Here,}{\text{initial temperature, T}}_{\text{i}}\text{= 27}\text{.0°C}\\ {\text{Diameter of hole at T}}_{\text{i}}{\text{, d}}_{\text{1}}\text{= 4}\text{.24 cm}\\ {\text{Final temperature, T}}_{\text{f}}\text{= 227°C}\\ \text{Let}{\text{diameter of hole at T}}_{\text{f}}{\text{= d}}_{\text{2}}\\ \text{Co-efficient of linear expansion of copper,}{\alpha }_{\text{Cu}}\text{= 1}{\text{.70 × 10}}^{\text{–5}}{\text{K}}^{\text{–1}}\\ \text{If the}\text{co-efficient of superficial expansion}\text{is β}\text{and}\\ \text{change}\text{in}\text{temperature is}\text{ΔT,}\text{we have,}\\ \frac{\text{Change}\text{in}\text{area}\left(\text{ΔA}\right)}{\text{Original}\text{area}\left(\text{A}\right)}\text{=}\text{βΔT}\\ \frac{\left(\pi \frac{{\text{d}}_{\text{2}}^{\text{2}}}{\text{4}}\text{–}\pi \frac{{\text{d}}_{\text{1}}^{\text{2}}}{\text{4}}\right)}{\pi \left(\text{π}\frac{{\text{d}}_{\text{1}}^{\text{2}}}{\text{4}}\right)}\text{=}\frac{\text{ΔA}}{\text{A}}\\ \therefore \frac{\text{ΔA}}{\text{A}}\text{=}\frac{{\text{d}}_{\text{2}}^{\text{2}}{\text{– d}}_{\text{1}}^{\text{2}}}{{\text{d}}_{\text{1}}^{\text{2}}}\\ \text{As,}\text{β = 2}\alpha \\ \because \frac{{\text{d}}_{\text{2}}^{\text{2}}{\text{– d}}_{\text{1}}^{\text{2}}}{{\text{d}}_{\text{1}}^{\text{2}}}\text{= 2}\alpha \text{ΔT}\\ \frac{{\text{d}}_{\text{2}}^{\text{2}}}{{\text{d}}_{\text{1}}^{\text{2}}}\text{-1 = 2}\alpha \left({\text{T}}_{\text{f}}{\text{– T}}_{\text{i}}\right)\\ \frac{{\text{d}}_{\text{2}}^{\text{2}}}{{\left(\text{4}\text{.24 m}\right)}^{\text{2}}}\text{= 2×1}{\text{.7×10}}^{\text{-5}}{\text{K}}^{\text{–1}}\left(\text{227 – 27}\right)\text{°C +1}\\ {\text{d}}_{\text{2}}^{\text{2}}\text{= 17}\text{.98 × 1}{\text{.0068 cm}}^2\text{= 18}{\text{.1 cm}}^2\\ \therefore {\text{d}}_2\text{= 4}\text{.2544}c\text{m}\\ {\text{Change in diameter = d}}_{\text{2}}{\text{– d}}_{\text{1}}\\ \text{= 4}\text{.2544 cm – 4}\text{.24 cm = 0}\text{.0144 cm}\\ \therefore \text{The}\text{diameter of}\text{the}\text{hole}\text{increases by 1}{\text{.44×10}}^{\text{–2}}\text{cm}\text{.}\end{array}

Q.9 A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0 × 10^–5 K^–1; Young’s modulus of brass = 0.91 × 10¹¹ Pa.

Ans.

\begin{array}{l}\text{Here,}{\text{initial temperature,T}}_{\text{i}}\text{= 27°C}\\ {\text{Length of brass wire at T}}_{\text{i}}\text{,}\text{l = 1}\text{.8 m}\\ \text{Final temperature,}{\text{T}}_{\text{f}}\text{= –39°C}\\ \text{Diameter of wire,}\text{d = 2}\text{.0 mm = 2}\text{×}{\text{10}}^{\text{–3}}\text{m}\\ \text{Coefficient of linear expansion of brass,}\alpha \text{=}\text{2}\text{.0}\text{×}{\text{10}}^{\text{–5}}{\text{K}}^{\text{–1}}\\ \text{Young’s modulus for brass,}\text{Y}\text{=}\text{0}\text{.91}\text{×}{\text{10}}^{\text{11}}\text{Pa}\\ \text{Let}\text{cross-sectional}\text{area}\text{of}\text{brass}\text{wire = A}\\ \text{Let}\text{tension produced in the brass}\text{wire = F}\\ \text{Young’s modulus is given as,}\text{Y}\text{=}\frac{\text{Stress}}{\text{Strain}}\\ \text{Y=}\frac{\text{F/A}}{\text{ΔL}/L}\\ \therefore \text{ΔL =}\frac{\text{F × L}}{\text{A × Y}}\text{(i)}\\ \text{Here,}\\ \text{ΔL}\text{=}\text{Change in}\text{the length}\text{of}\text{wire,}\\ \text{Change in}\text{the length}\text{of}\text{wire}\text{is}\text{given by}\text{the}\text{relation,}\\ \text{ΔL =}\alpha \text{L}\left({\text{T}}_{\text{2}}{\text{–T}}_{\text{1}}\right)\left(\text{ii}\right)\\ \text{Equating equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we obtain:}\\ \alpha \text{L}\left({\text{T}}_{\text{2}}{\text{–T}}_{\text{1}}\right)\text{=}\frac{\text{FL}}{\pi {\left(\frac{\text{d}}{\text{2}}\right)}^{\text{2}}\text{×Y}}\\ \text{F =}\alpha \left({\text{T}}_{\text{2}}{\text{–T}}_{\text{1}}\right)\pi \text{Y}{\left(\frac{\text{d}}{\text{2}}\right)}^{\text{2}}\\ \text{F = 2}{\text{.0 × 10}}^{\text{–5}}{\text{K}}^{\text{–1}}\text{×}\left(\text{-39 – 27}\right)\text{°C × 3}\text{.14 × 0}{\text{.91×10}}^{\text{11}}\text{Pa ×}{\left(\frac{\text{2}{\text{.0 × 10}}^{\text{–3}}\text{m}}{\text{2}}\right)}^{\text{2}}\\ \text{F = -3}{\text{.8 ×10}}^{\text{2}}\text{N}\\ \text{(The negative sign shows that the tension is directed inward}\text{.)}\\ \therefore \text{The tension produced in the wire is 3}{\text{.8 ×10}}^{\text{2}}\text{N}\text{.}\end{array}

Q.10 A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10^–5 K^–1, steel = 1.2 × 10^-5 K^-1).

Ans.

\begin{array}{l}\text{Here,}\text{initial temperature,}{\text{T}}_{\text{i}}\text{=}\text{40°C}\\ {\text{Final temperature,T}}_{\text{f}}\text{=}\text{250°C}\\ \text{Change in temperature, ΔT}\text{=}{\text{T}}_{\text{f}}\text{–}{\text{T}}_{\text{i}}\text{=}\text{210°C}\\ \text{Length of brass rod at}{\text{T}}_{\text{i}}{\text{, l}}_{\text{i}}\text{= 50 cm}\\ \text{Diameter of brass rod at}{\text{T}}_{\text{i}}{\text{, d}}_{\text{1}}\text{= 3}\text{.0 mm}\\ \text{Length of steel rod at}{\text{T}}_{\text{f}}{\text{, l}}_{\text{2}}\text{=}\text{50 cm}\\ \text{Diameter of steel rod at}{\text{T}}_{\text{f}}{\text{, d}}_{\text{2}}\text{= 3}\text{.0 mm}\\ \text{Coefficient of linear expansion of the}\text{brass,}{\alpha }_{\text{b}}\text{= 2}\text{.0}\text{×}{\text{10}}^{\text{–5}}{\text{K}}^{\text{–1}}\\ \text{Coefficient of linear expansion of the}\text{steel,}{\alpha }_{\text{s}}\text{= 1}{\text{.2 × 10}}^{\text{–5}}{\text{K}}^{\text{–1}}\\ \text{For expansion in the brass rod, we have:}\\ \frac{\text{Change}\text{in}\text{length}\left({\text{Δl}}_{\text{1}}\right)}{\text{Original}\text{length}\left({\text{l}}_{\text{1}}\right)}\text{=}{\alpha }_{\text{b}}\text{ΔT}\\ \therefore {\text{Δl}}_{\text{1}}\text{=}\text{50 cm}\text{×}\left(\text{2}\text{.1}\text{×}{\text{10}}^{\text{–5}}{\text{K}}^{\text{–1}}\right)\text{×}\text{210°C}\\ {\text{Δl}}_{\text{1}}\text{=}\text{0}\text{.2205}\text{cm}\\ \text{For expansion in the steel rod, we have,}\\ \frac{\text{Change}\text{in}\text{length}\left({\text{Δl}}_{\text{2}}\right)}{\text{Original}\text{length}\left({\text{l}}_{\text{2}}\right)}\text{=}{\alpha }_{\text{s}}\text{ΔT}\\ \therefore {\text{Δl}}_{\text{2}}\text{=}\text{50 cm}\text{×}\left(\text{1}{\text{.2 × 10}}^{\text{–5}}{\text{K}}^{\text{–1}}\right)\text{×}\text{210°C}\\ {\text{Δl}}_{\text{2}}\text{= 0}\text{.126}\text{cm}\\ \text{Total change in the lengths of brass and steel}\text{is}\text{given}\text{as:}\\ {\text{Δl = Δl}}_{\text{1}}{\text{+ Δl}}_{\text{2}}\\ \text{Δl = 0}\text{.2205 + 0}\text{.126 = 0}\text{.346 cm}\\ \text{Total change in the length of the joined rod}\text{=}\text{0}\text{.346 cm}\\ \text{As the rod expands freely from both ends, no thermal}\text{stress}\\ \text{is produced at the junction}\text{.}\end{array}

Q.11 The coefficient of volume expansion of glycerin is 49 × 10^–5 K^–1. What is the fractional change in its density for a 30°C rise in temperature?

Ans.

\begin{array}{l}\text{Here,}\text{coefficient of volume expansion of glycerin,}{\alpha }_{\text{V}}\text{=}{\text{49 × 10}}^{\text{–5}}{\text{K}}^{\text{–1}}\\ \text{Increase in the}\text{temperature, ΔT = 30°C}\\ \text{Fractional change in volume of}\text{glycerin}\text{=}\frac{\text{ΔV}}{\text{V}}\\ \text{The}\text{change in}\text{volume}\text{is related to the change in}\text{temperature as:}\\ \frac{\text{ΔV}}{\text{V}}\text{=}{\alpha }_{\text{V}}\text{ΔT}\\ {\text{V}}_{{\text{T}}_{\text{2}}}{\text{– V}}_{{\text{T}}_{\text{1}}}{\text{= V}}_{{\text{T}}_{\text{1}}}{\alpha }_{\text{V}}\text{ΔT}\\ \frac{\text{m}}{{\text{ρ}}_{{\text{T}}_{\text{2}}}}\text{–}\frac{\text{m}}{{\text{ρ}}_{{\text{T}}_{\text{1}}}}\text{=}\frac{\text{m}}{{\text{ρ}}_{{\text{T}}_{\text{1}}}}{\alpha }_{\text{V}}\text{ΔT}\\ \text{Here, m = Mass of glycerin}\\ {\text{ρ}}_{{\text{T}}_{\text{1}}}\text{= Initial density at}{\text{T}}_{\text{1}}\\ {\text{ρ}}_{{\text{T}}_{\text{2}}}{\text{= Final density at T}}_{\text{2}}\\ \frac{{\text{ρ}}_{{\text{T}}_{\text{1}}}{\text{– ρ}}_{{\text{T}}_{\text{2}}}}{{\text{ρ}}_{{\text{T}}_{\text{2}}}}\text{=}{\alpha }_{\text{V}}\text{ΔT}\\ \text{Here,}\frac{{\text{ρ}}_{{\text{T}}_{\text{1}}}{\text{– ρ}}_{{\text{T}}_{\text{2}}}}{{\text{ρ}}_{{\text{T}}_{\text{2}}}}\text{=Fractional change in the}\text{density}\\ \therefore \text{Fractional change in density of}{\text{glycerin = 49×10}}^{\text{–5}}{\text{K}}^{\text{–1}}\text{× 30°C}\\ \text{=1}{\text{.47×10}}^{\text{–2}}\end{array}

Q.12 A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g^–1K^–1.

Ans.

\begin{array}{l}\text{Here,}{\text{power of drilling machine, P = 10 kW = 10 × 10}}^{\text{3}}\text{W}\\ \text{Mass of aluminum block, m = 8}{\text{.0 kg = 8 × 10}}^{\text{3}}\text{g}\\ \text{Time for which the machine is used, t = 2}\text{.5 min}\\ \text{t = 2}\text{.5 × 60}\text{s =150 s}\\ \text{Specific heat of aluminium, c = 0}{\text{.91 J g}}^{\text{–1}}{\text{K}}^{\text{–1}}\\ \text{Let}\text{increase in the temperature = ΔT}\\ \text{Total energy of the}\text{drilling machine = Pt}\\ {\text{Pt =10×10}}^{\text{3}}\text{W × 150 s = 1}{\text{.5 × 10}}^{\text{6}}\text{J}\\ \text{As only 50\% of energy is useful,}\\ \therefore \text{Useful}\text{energy,}\text{ΔQ =}\frac{\text{1}}{\text{2}}\text{× 1}{\text{.5 × 10}}^{\text{6}}\text{J}\\ \text{ΔQ = 7}{\text{.5 × 10}}^{\text{5}}\text{J}\\ \text{As}\text{ΔQ = mcΔT}\\ \therefore \text{ΔT =}\frac{\text{ΔQ}}{\text{mc}}\\ \text{ΔT =}\frac{\text{7}{\text{.5 × 10}}^{\text{5}}\text{J}}{{\text{8×10}}^{\text{3}}\text{g × 0}{\text{.91 J g}}^{\text{–1}}{\text{K}}^{\text{–1}}}\\ \text{ΔT = 103°C}\\ \therefore \text{Increase in the temperature of the block = 103°C}\end{array}

Q.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt?

(Specific heat of copper = 0.39 Jg^–1K^–1; heat of fusion of water = 335 J g^–1).

Ans.

\begin{array}{l}\text{Here,}\text{mass of copper block, m = 2}\text{.5 kg = 2500 g}\\ \text{Increase in temperature of the copper block, Δθ = 500°C}\\ \text{Specific heat of copper, C = 0}{\text{.39 J g}}^{\text{–1}}{\text{°C}}^{\text{–1}}\\ {\text{Heat of fusion of water, L = 335 J g}}^{\text{–1}}\\ \text{Heat lost}\text{by}\text{the copper block}\text{is}\text{given}\text{as: Q = mCΔθ}\\ \text{Q = 2500 g × 0}{\text{.39 J g}}^{\text{–1}}{\text{°C}}^{\text{–1}}\text{× 500°C = 487500 J}\\ \text{Let}\text{mass}\text{of}\text{ice}\text{melted}\text{=}\text{m’}\\ \text{Heat attained by the melted ice, Q = m’L}\\ \text{As,}\text{Heat attained by ice}\text{=}\text{Heat lost by copper}\\ \therefore \text{m’L = 487500 J}\\ \therefore \text{m’ =}\frac{\text{487500}}{\text{L}}\text{=}\frac{\text{487500 J}}{{\text{335 J g}}^{\text{–1}}}\text{= 1455}\text{.22}\text{g}\\ \text{m’ =1}\text{.45}\text{kg}\\ \therefore \text{Maximum amount of ice that can melt = 1}\text{.45 kg}\end{array}

Q.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

Ans.

Here, mass of metal, m = 0.20 kg = 200 g

Initial temperature of metal, T₁ = 150°C

Final temperature of metal, T₂ = 40°C

Decrease in temperature of the metal: ΔT = T₁ – T₂ = 150°C – 40°C = 110°C

Water equivalent of calorimeter, w = 0.025 kg = 25 g

Volume of the water, V = 150 cm³

Mass (M) of the water at 27°C: 150 g × 1 = 150 g

Specific heat of the water, C_w = 4.186 Jg^-1K

Let specific heat of metal = C

$\begin{array}{l}\text{Heat lost by the metal}\text{=}\text{Heat attained by the water}\text{and colorimeter system}\hfill \\ \therefore {\text{mCΔT = (M+w)C}}_{\text{w}}\text{ΔT’}\hfill \\ \text{200kg × C × 110°C = (150 + 25)kg × 4}{\text{.186 Jg}}^{\text{-1}}\text{K × 13°C}\hfill \\ \therefore \text{C =}\frac{\text{175 kg × 4}{\text{.186 Jg}}^{\text{-1}}\text{K × 13°C}}{\text{110 °C × 200 kg}}\hfill \\ \text{C = 0}\text{.43}{\text{Jg}}^{\text{-1}}{\text{K}}^{\text{-1}}\hfill \\ \text{If}\text{some}\text{heat}\text{is}\text{lost}\text{to}\text{the}\text{surroundings,}\text{then}\text{the}\text{value}\text{of}\hfill \\ \text{C}\text{will}\text{be}\text{less}\text{than}\text{the}\text{actual}\text{value}\text{of}\text{C}\text{.}\hfill \end{array}$

Q.15 Given below are observations on molar specific heats at room temperature of some common gases.

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Ans.

The gases tabulated in the given table are diatomic in nature. In addition to the translational degree of freedom, they have other degrees of freedom (modes of motion).

Heat must be supplied to increase the temperature of the given gases. The average energy of all the modes of motion increases with it. Thus, the molar specific heat of diatomic gas is greater than that of monatomic gas. If we consider only the rotational mode of motion, then

\begin{array}{l}{\text{Molar specific heat of a diatomic gas,C}}_{\text{v}}\text{=}\frac{\text{5}}{\text{2}}\text{R =}\frac{\text{5}}{\text{2}}\text{× 1}\text{.98}\\ {\text{C}}_{\text{v}}\text{= 4}\text{.95}{\text{cal mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\\ \text{Except}\text{chlorine, all the observations in the given table}\\ \text{agree}\text{fairly}\text{well}\text{with}\left(\frac{\text{5}}{\text{2}}\text{R}\right)\text{.}\end{array}

This is because at room temperature, chlorine also has vibrational modes of motion in addition to rotational and translational modes of motion.

Q.16 Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂ co-exist in equilibrium?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO₂?
(c) What are the critical temperature and pressure for CO₂? What is their significance?
(d) Is CO₂ solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm?

Ans.

(a) The P-T phase diagram for CO₂ is shown in the given figure. In the given phase diagram, C is the triple point of the CO₂. It implies that the solid, liquid, and vapour phases of CO₂ co-exist in equilibrium at the temperature and pressure corresponding to this point (i.e., at temperature = –56.6 °C and pressure = 5.11 atm).

(b) Both the fusion and boiling points of CO₂ fall with the decrease in pressure.

(c) In case of CO₂, the critical temperature and critical pressure are 31.1°C and 73 atm respectively. If the temperature of CO₂ is greater than 31.1°C, it cannot be liquefied, howsoever large pressure we may apply.

(d) From the P-T phase diagram of CO₂, it can be concluded that,

(a) At –70°C, under 1 atm pressure CO₂ is a vapour.

(b) At –60°C, under 10 atm pressure CO₂ is a solid.

(c) At 15°C, under 56 atm pressure CO₂ is a liquid.

Q.17 Answer the following questions based on the P–T phase diagram of CO₂:
(a) CO₂ at 1 atm pressure and temperature –60°C is compressed isothermally. Does it go through a liquid phase?
(b) What happens when CO₂ at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO₂ at 10 atm pressure and temperature – 65°C as it is heated up to room temperature at constant pressure.
(d) CO₂ is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?

Ans.

(a) The P-T phase diagram for CO₂ is shown in the given figure. In the given phase diagram, at 1 atm pressure and at –60°C temperature, CO₂ lies to the left of -56.6°C (triple point C), therefore, it lies in the region of vapour and solid phase. Therefore, CO₂ condenses directly into the solid state, without becoming liquid.

(b) Since at the 4 atm pressure CO₂ lies below 5.11 atm (triple point C), therefore, CO₂ lies in the region of vapour and solid phase. Hence, CO₂ condenses into the solid state directly, without becoming liquid.

(c) When solid CO₂ (at 10 atm pressure and at–65°C) is heated, it changes to the liquid phase and then to the vapour phase. At pressure=10 atm if a horizontal line is drawn parallel to the temperature axis, then the fusion and boiling points of CO₂ are given by the intersection point where this parallel line crosses the fusion and vaporisation curves.

(d) If CO₂ is heated to 70°C and compressed isothermally, then it will not be converted to the liquid state. This is because; the critical temperature of CO₂ is lower than 70°C. It will exist in the vapour state, but will depart more and more from its ideal behaviour with the increase in pressure.

Q.18 A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g^–1.

Ans.

\begin{array}{l}\text{Here,}\text{Initial temperature}\text{of}\text{the}{\text{body, T}}_{\text{1}}\text{=}\text{101°F}\\ {\text{Final temperature of the body,T}}_{\text{2}}\text{=}\text{98°F}\\ \text{Change in temperature}\text{of}\text{body, ΔT =}\left[\left(\text{101°F – 98 °F}\right)\text{×}\frac{\text{5}}{\text{9}}\right]\text{°C}\\ \text{Time taken to decrease}\text{the temperature,}\text{t}\text{=}\text{20 min}\\ \text{Mass of child, m}\text{=}\text{30 kg = 30}\text{×}{\text{10}}^{\text{3}}\text{g}\\ \text{Specific heat of human body}\text{=}\text{Specific heat of}\text{water}\text{=}\text{c}\\ {\text{c = 1000 cal kg}}^{\text{-1}}{\text{°C}}^{\text{-1}}\\ \text{Latent heat of evaporation of the}{\text{water,L = 580 calg}}^{\text{–1}}\\ \text{Heat lost by the child is given as,}\text{Δθ}\text{=}\text{mcΔT}\\ {\text{Δθ =30 kg × 1000 cal kg}}^{\text{-1}}{\text{°C}}^{\text{-1}}\text{×}\left(\text{101 – 98}\right)\text{°F ×}\frac{\text{5}}{\text{9}}\\ \text{Δθ}\text{= 50000}\text{cal}\\ \text{Consider mass of the water evaporated from}\text{the body}\\ \text{of}\text{the}\text{child in 20 min}\text{=}{\text{m}}_{\text{1}}\\ \text{Heat loss}\text{through water is given as,}\text{Δθ}\text{=}{\text{m}}_{\text{1}}\text{L}\\ \therefore {\text{m}}_{\text{1}}\text{=}\frac{\text{Δθ}}{\text{L}}\\ {\text{m}}_{\text{1}}\text{=}\frac{\text{50000}\text{cal}}{{\text{580 calg}}^{\text{–1}}}\\ {\text{m}}_{\text{1}}\text{= 86}\text{.2}\text{g}\\ \therefore \text{Average rate of extra evaporation due}\text{to the}\text{drug =}\frac{{\text{m}}_{\text{1}}}{\text{t}}\\ \frac{{\text{m}}_{\text{1}}}{\text{t}}\text{=}\frac{\text{86}\text{.2 g}}{\text{20 min}}\text{= 4}\text{.31}{\text{g min}}^{\text{-1}}\end{array}

Q.19 A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C, and co-efficient of thermal conductivity of thermacole is 0.01 J s^–1m^–1K^–1. [Heat of fusion of water = 335 × 10³ J kg^–1]

Ans.

\begin{array}{l}\text{Here,}\text{side of the given cubical ice box, s = 30 cm = 0}\text{.3 m}\\ \text{Thickness of ice box, l = 5}\text{.0 cm = 0}\text{.05 m}\\ \text{Mass of the}\text{ice placed in the ice box, m}\text{=}\text{4 kg}\\ \text{Time difference, t}\text{=}\text{6 h = 6 × 60 × 60 s}\\ \text{Outside temperature,}\text{T = 45°C}\\ \text{Coefficient of thermal conductivity of the}\text{thermacole,}\\ \text{K = 0}{\text{.01 J s}}^{\text{–1}}{\text{m}}^{\text{–1}}{\text{K}}^{\text{–1}}\\ \text{Latent}\text{heat}\text{of}\text{fusion,}\text{L}\text{=}\text{335}\text{×}{\text{10}}^{\text{3}}{\text{Jkg}}^{\text{–1}}\\ \text{Let total amount of ice that melts in 6 h}\text{=}\text{m’}\\ \text{Heat lost by the food}\text{is}\text{given}\text{as:}\text{θ}\text{=}\frac{\text{KA}\left(\text{T-0}\right)\text{t}}{\text{l}}\\ \text{Here,}{\text{A = Surface area of the box = 6s}}^{\text{2}}\\ \text{A = 6×}{\left(\text{0}\text{.3 m}\right)}^{\text{2}}\text{= 0}{\text{.54 m}}^{\text{3}}\\ \therefore \text{θ =}\frac{\text{0}\text{.01 × 0}{\text{.54 m}}^{\text{3}}\text{×}\left(\text{45°C}-0\text{°C}\right)\text{× 6 × 60 × 60 s}}{\text{0}\text{.05 m}}\\ \text{θ = 104976}\text{J}\\ \text{As}\text{θ = m’L}\\ \therefore \text{m’ =}\frac{\text{θ}}{\text{L}}\text{=}\frac{\text{104976 J}}{{\text{335 × 10}}^{\text{3}}{\text{Jkg}}^{\text{–1}}}\\ \text{m’ = 0}\text{.313}\text{kg}\\ \text{Mass of the}\text{ice remained}\text{=}(\text{4 – 0}\text{.313) kg = 3}\text{.687 kg}\\ \therefore \text{The}\text{amount of ice left after 6 h is 3}\text{.687 kg}\text{.}\end{array}

Q.20 A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg min^-1 when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s^–1m^–1K^–1; Heat of vaporisation of water = 2256 × 10³ Jkg^–1.

Ans.

\begin{array}{l}\text{Here,}\text{base area of the brass}\text{boiler, A}\text{=}\text{0}{\text{.15 m}}^{\text{2}}\\ \text{Thickness of the}\text{brass}\text{boiler, l = 1}\text{.0 cm = 0}\text{.01 m}\\ \text{Rate of}\text{boiling of}\text{water, R = 6}{\text{.0 kg min}}^{\text{-1}}\\ \text{Mass, m = 6 kg}\\ \text{Time, t = 1 min = 60 s}\\ \text{Thermal conductivity of the}{\text{brass, K = 109 J s}}^{\text{–1}}{\text{m}}^{\text{–1}}{\text{K}}^{\text{–1}}\\ \text{Heat of vaporisation}\text{of}\text{water, L = 2256}\text{×}{\text{10}}^{\text{3}}{\text{J kg}}^{\text{–1}}\\ \text{Heat flowing into water through the brass base of the}\\ \text{boiler,}\text{θ}\text{=}\frac{\text{KA}\left({\text{T}}_{\text{1}}{\text{-T}}_{\text{2}}\right)\text{t}}{\text{l}}\text{(i)}\\ {\text{Here, T}}_{\text{1}}\text{and}{\text{T}}_{\text{2}}\text{represent temperature of the flame in}\\ \text{contact with boiler}\text{and boiling point of water}\left(\text{100°C}\right)\\ \text{respectively}\text{.}\\ \text{Heat required for boiling the water}\text{is}\text{given}\text{as:}\\ \text{θ = mL}\left(\text{ii}\right)\\ \text{Equating equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we obtain:}\\ \text{mL =}\frac{\text{KA}\left({\text{T}}_{\text{1}}{\text{– T}}_{\text{2}}\right)\text{t}}{\text{l}}\\ \therefore {\text{T}}_{\text{1}}{\text{– T}}_{\text{2}}\text{=}\frac{\text{mLl}}{\text{KAt}}\\ {\text{T}}_{\text{1}}{\text{– T}}_{\text{2}}\text{=}\frac{\text{6 kg}\times {\text{2256 × 10}}^{\text{3}}{\text{J kg}}^{\text{–1}}\text{× 0}\text{.01}}{{\text{109 Js}}^{\text{–1}}{\text{m}}^{\text{–1}}{\text{K}}^{\text{–1}}\text{× 0}{\text{.15 m}}^{\text{2}}\text{× 60 s}}\\ {\text{T}}_{\text{1}}{\text{– T}}_{\text{2}}\text{= 137}\text{.98°C}\\ \therefore \text{The}\text{temperature of the part of the flame in}\text{contact}\\ \text{with}\text{boiler = 237}\text{.98°C}\end{array}

Q.21 Explain why:
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

Ans.

(a) This is because a body with a large reflectivity is a poor absorber of light radiations and poor absorbers are poor emitters. Thus, a body with a large reflectivity is a poor emitter.

(b) Since brass is a good conductor of heat , when we touch a brass tumbler, heat is conducted from our body to the brass tumbler. Therefore, the temperature of our body decreases and we feel cooler.

On the contrary, wood is a poor conductor of heat.Thus, there is only a negligible fall in the temperature of our body and we feel comparatively less cool.

Hence, a brass tumbler feels much colder than a wooden tray on a chilly day.

(c) An optical pyrometer calibrated for an ideal black body radiation gives a very low value for temperature of a red hot iron piece placed in the open.

Black body radiation equation is given as:

\text{E}\text{=}\text{σ}\left({\text{T}}^{\text{4}}\text{–}{\text{T}}_{\text{0}}^{\text{4}}\right)

Here, E represents the energy radiation

T and T_o represent temperature of the optical pyrometer and Temperature of the open space respectively.

σ is a constant

Thus, a rise in the temperature of open space lowers the radiation energy.

When the same piece of iron is kept in a furnace, its radiation energy is given as: E = σT⁴

(d) The atmospheric gases reflect infrared radiations from Earth back to its surface. Hence, the heat radiations collected by Earth from the Sun during day are kept trapped by the atmosphere.. All the heat would be radiated back from earth’s surface without atmospheric gases and the Earth would become too cold to live.

(e) The heating system based on the circulation of steam is more useful in warming a building than that based on the circulation of hot water. This is due to the fact that steam contains excessive heat in the form of latent heat which is 540 cal/g.

Q.22 A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.

Ans.

\begin{array}{l}\text{Here, temperature of the surroundings}\text{=}{\text{T}}_{\text{0}}\text{=}\text{20°C}\\ \text{Let temperature of the body}\text{=}\text{T}\\ \text{According to Newton’s law of cooling, we have}\\ \frac{\text{dT}}{\text{dt}}\text{= -K}\left({\text{T – T}}_{\text{0}}\right)\\ \therefore \frac{\text{dT}}{{\text{T – T}}_{\text{0}}}\text{= -Kdt}\\ \text{Here,}\text{K}\text{is}\text{a}\text{constant}\text{.}\\ \text{Let}\text{temperature}\text{of}\text{the}\text{body}\text{falls}\text{from}{\text{T}}_{\text{i}}\text{to}{\text{T}}_{\text{f}}\text{in}\text{time}\text{t,}\\ \text{then}\text{integrating}\text{the}\text{above}\text{relation,}\text{we}\text{have}\\ {\displaystyle \underset{{\text{T}}_{\text{i}}}{\overset{{\text{T}}_{\text{f}}}{\int }}\frac{\text{dT}}{{\text{T-T}}_{\text{0}}}\text{= –}{\displaystyle \underset{\text{0}}{\overset{\text{t}}{\int }}\text{Kdt}}}\\ {\left[{\text{log}}_{\text{e}}\left({\text{T-T}}_{\text{0}}\right)\right]}_{{\text{T}}_{\text{i}}}^{{\text{T}}_{\text{f}}}\text{= -Kt}\\ {\text{log}}_{\text{e}}\frac{{\text{T}}_{\text{f}}{\text{– T}}_{\text{0}}}{{\text{T}}_{\text{i}}{\text{– T}}_{\text{0}}}\text{= -Kt}\\ \text{2}{\text{.3026 log}}_{\text{10}}\frac{{\text{T}}_{\text{f}}{\text{– T}}_{\text{0}}}{{\text{T}}_{\text{i}}{\text{– T}}_{\text{0}}}\text{= -Kt}\\ \text{2}{\text{.3026 log}}_{\text{10}}\frac{{\text{T}}_{\text{i}}{\text{– T}}_{\text{0}}}{{\text{T}}_{\text{f}}{\text{– T}}_{\text{0}}}\text{= Kt}\\ \text{t =}\frac{\text{2}\text{.3026}}{\text{K}}{\text{log}}_{\text{10}}\frac{{\text{T}}_{\text{1}}{\text{– T}}_{\text{0}}}{{\text{T}}_{\text{2}}{\text{– T}}_{\text{0}}}\text{(A)}\\ \text{According}\text{to}\text{the}\text{first}\text{condition}\text{of}\text{the}\text{problem,}\text{we}\text{have}\\ {\text{T}}_{\text{i}}\text{= 80°C,}{\text{T}}_{\text{f}}\text{= 50°C,}\\ {\text{T}}_{\text{0}}\text{= 20°C,}\text{t = 5}\text{min = 5 × 60}\text{s}\\ \text{Substituting}\text{these}\text{values}\text{in}\text{equation}\text{(A),}\text{we}\text{obtain:}\\ \text{5 × 60}\text{s =}\frac{\text{2}\text{.3026}}{\text{K}}{\text{log}}_{\text{10}}\frac{\text{80°}C\text{– 20°}C}{\text{50°}C\text{– 20°}C}\\ \text{5 × 60}\text{s =}\frac{\text{2}\text{.3026}}{\text{K}}{\text{log}}_{\text{10}}\text{2 (i)}\\ \text{According}\text{to}\text{the}\text{second}\text{condition}\text{of}\text{the}\text{problem,}\\ {\text{T}}_{\text{i}}\text{= 60°C,}{\text{T}}_{\text{f}}\text{= 30°C,}\\ {\text{T}}_{\text{0}}\text{= 20°C,}\text{t = ?}\\ \text{Substituting}\text{these}\text{values}\text{in}\text{equation}\text{(A),}\text{we}\text{obtain:}\\ \text{t =}\frac{\text{2}\text{.3026}}{\text{K}}{\text{log}}_{\text{10}}\frac{\text{60°C – 20°C}}{\text{30°C – 20°C}}\\ \text{t =}\frac{\text{2}\text{.3026}}{\text{K}}{\text{log}}_{\text{10}}\text{4 (ii)}\\ \text{Dividing}\text{equation}\text{(ii)}\text{by}\text{equation}\text{(i),}\text{we}\text{obtain:}\\ \frac{\text{t}}{\text{5 × 60 s}}\text{=}\frac{{\text{log}}_{\text{10}}\text{4}}{{\text{log}}_{\text{10}}\text{2}}\\ \frac{\text{t}}{\text{5 × 60 s}}\text{=}\frac{\text{0}\text{.6021}}{\text{0}\text{.3010}}\text{= 2}\\ \therefore \text{t = 600}\text{s = 10}\text{min}\\ \therefore \text{Time}\text{taken}\text{to}\text{cool}\text{the}\text{body}\text{from}\text{60°C}\text{to}\text{30°C = 10}\text{minutes}\end{array}