Question

A thermally insulated piece of metal is heated under atmospheric pressure by an electric current so that it receives electric energy at a constant power P. This leads to an increase of absolute temperature T of the metal with time t as follows:
$T(t)={{T}_{0}}{{[1+a(t-{{t}_{0}})]}^{0.25}}$. Here $a$, ${{t}_{0}}$ and ${{T}_{0}}$ are constants. The heat capacity ${{C}_{p}}(T)$ of the metal is
A. $\dfrac{4P}{a{{T}_{0}}}$
B. $\dfrac{4P{{T}^{3}}}{aT_{0}^{4}}$
C. $\dfrac{2P{{T}^{3}}}{aT_{0}^{4}}$
D. $\dfrac{2P}{a{{T}_{0}}}$

Answer 1

Hint: Heat capacity equation, $C=\dfrac{dQ}{dt}$ and Change of enthalpy, $dH=P.dt$ will lead to the answer. Heat can be written as the product of power and time. Here, the change of enthalpy will be equal to the amount of heat at isobaric condition.

Complete step by step answer:
Heat capacity can be written as,
$C=\dfrac{dQ}{dt}$…………..(1) It represents the quantity of heat transferred per unit time and it is valid only if the phase transitions are absent.
According to the enthalpy of isobaric change of state, the change of enthalpy will be equal to the amount of heat at isobaric condition.
$\dfrac{dQ}{dt}=dH=m{{c}_{p}}dT$, where $\text{dH}$ is the change of enthalpy, $m$ is the mass, ${{c}_{p}}$ is the isobaric heat capacity and $dT$ is the change of temperature.
$\text{Heat = power }\times \text{ time}$
$dH=P.dt$
$m{{c}_{p}}dT=P.dt$……………….(2)
So, the heat capacity will be,
$m{{c}_{p}}=\dfrac{P.dt}{dT}$, $m{{c}_{p}}$ collectively known as heat capacity $({{C}_{p}})$.
${{C}_{p}}=\dfrac{P.dt}{dT}$……………(3)
We can find the $\dfrac{dt}{dT}$ from the question.
Since $T(t)={{T}_{0}}{{[1+a(t-{{t}_{0}})]}^{0.25}}$, where $a$, ${{t}_{0}}$ and ${{T}_{0}}$ are constants.
It can simplify to make the calculation as simple by taking the fourth power of L.H.S and R.H.S.
$\Rightarrow T_{0}^{4}[1+a(t-{{t}_{0}})]={{T}^{4}}$
$\Rightarrow 4{{T}^{3}}dT=aT_{0}^{4}dt$
Thus, after differentiation, we will get
$\Rightarrow \dfrac{dt}{dT}=\dfrac{4{{T}^{3}}}{aT_{0}^{4}}$
We can assign this into the equation 3
${{C}_{p}}=\dfrac{P.dt}{dT}$
Therefore, the heat capacity of the metal is,
${{C}_{p}}=\dfrac{P\times 4{{T}^{3}}}{aT_{0}^{4}}$
Hence the correct option is B.

Note: If you are converting this equation, $T(t)={{T}_{0}}{{[1+a(t-{{t}_{0}})]}^{0.25}}$ to $T_{0}^{4}[1+a(t-{{t}_{0}})]={{T}^{4}}$, you should not forgot to put power to the LHS also. Otherwise the answer will be wrong. This step will be useful to make the calculation not complex.