Question

A thermally insulated piece of metal is heated under atmosphere by an electric current so that it receives electric energy at a constant power $P$. This leads to an increase of the absolute temperature $T$ of the metal with time $t$ as follows $T=a{{t}^{1/4}}$ Then the heat capacity $C_{p}$ is :
(A) $\dfrac{4P{{T}^{3}}}{{{a}^{4}}}$
(B) $\dfrac{4P{{T}^{2}}}{{{a}^{3}}}$
(C) $4P{{T}^{2}}$
(D) none of these

Answer 1

Hint
We know that electrical energy is derived from the electric potential energy or the kinetic energy. When we consider it loosely then electrical energy refers to the energy that has been converted from the electric potential energy. Based on this question we have to solve this question.

Complete step by step answer
We know that heat will be equal to the product of power and time.
$\therefore d H=P d t \quad $
$\therefore C_{p}=\dfrac{P d t}{d T}=\left(\dfrac{P}{{d T} /{d t}}\right)$
Here, Increase in temperature,
$T=a t^{\dfrac{1}{4}}$
$\Rightarrow \dfrac{T}{a}={{t}^{\dfrac{1}{4}}}$
$\therefore t=\left(\dfrac{T^{4}}{a^{4}}\right)$
After differentiating both sides, we get,
$\therefore \dfrac{dT}{dt}=\dfrac{a}{4}{{\left( \dfrac{T}{a} \right)}^{-3}}$
$=\dfrac{a}{4}{{\left( \dfrac{T}{a} \right)}^{-3}}$
$=\dfrac{{{a}^{4}}}{4{{T}^{3}}}$
Substituting the equations obtained, we get,
$\Rightarrow {{C}_{p}}=\dfrac{P}{{dT}/{dt}}$
$\therefore \dfrac{d T}{d t} = \dfrac{a}{4}\left(\dfrac{T}{a}\right)^{-3} = \dfrac{a}{4}\left(\dfrac{T}{a}\right)^{-3} = \dfrac{a^{4}}{4 T^{3}} $
$\Rightarrow C_{p}=\dfrac{P}{{d T} /{d t}}=\dfrac{4 P T^{3}}{a^{4}}$
Therefore, the correct answer is Option (A).

Note
We know that the absolute temperature is defined as the temperature of an object on a scale where $0$ is taken as the absolute zero. The absolute zero at which a thermodynamic system is considered to be the lowest possible energy.