A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as : $$\mathrm{T}(\mathrm{t})=\mathrm{T}_{0}\left(1+\beta \mathrm{t}^{1 / 4}\right)$$ where $\beta$ is a constant with appropriate dimension while $\mathrm{T}_{0}$ is a constant with dimension of temperature. The heat capacity of the metal is :

(A) $\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_{0}\right)^{3}}{\beta^{4} \mathrm{~T}_{0}^{4}}$

(B) $\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_{0}\right)}{\beta^{4} \mathrm{~T}_{0}^{2}}$

(C) $\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_{0}\right)^{4}}{\beta^{4} \mathrm{~T}_{0}^{5}}$

(D) $\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_{0}\right)^{2}}{\beta^{4} \mathrm{~T}_{0}^{3}}$

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Heat & Thermodynamics

A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as :

T (t) = T_{0} (1 + β t^{1/4})

where

β

is a constant with appropriate dimension while

T_{0}

is a constant with dimension of temperature. The heat capacity of the metal is :

(A) $\frac{4 P ( T ( t ) - T _{0} ) ^{3}}{β ^{4} T _{0}^{4}}$ ✓ Correct
(B) $\frac{4 P ( T ( t ) - T _{0} )}{β ^{4} T _{0}^{2}}$
(C) $\frac{4 P ( T ( t ) - T _{0} ) ^{4}}{β ^{4} T _{0}^{5}}$
(D) $\frac{4 P ( T ( t ) - T _{0} ) ^{2}}{β ^{4} T _{0}^{3}}$

Answer: (A)

Solution

The heat capacity relates power to the rate of temperature change. Starting with

P = \frac{d Q}{d t} = S \frac{d T}{d t}

, where

S

is the heat capacity.

Differentiate the given temperature function:

\frac{d T}{d t} = T_{0} β \cdot \frac{1}{4} t^{- 3/4} = \frac{β T _{0}}{4} t^{- 3/4}

Therefore:

S = \frac{P}{d T / d t} = \frac{4 P}{β T _{0}} t^{3/4}

From

T (t) = T_{0} (1 + β t^{1/4})

, we get:

β t^{1/4} = \frac{T ( t ) - T _{0}}{T _{0}}

t^{3/4} = (\frac{T ( t ) - T _{0}}{β T _{0}})^{3}

Substituting back:

S = \frac{4 P}{β T _{0}} (\frac{T ( t ) - T _{0}}{β T _{0}})^{3} = \frac{4 P ( T ( t ) - T _{0} ) ^{3}}{β ^{4} T _{0}^{4}}

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Single correct option

Practise more: all JEE Advanced PYQs on Heat & Thermodynamics →

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Heat & Thermodynamics

Solution

More JEE Advanced questions on Heat & Thermodynamics