Question

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 $T\left(t\right)=T_0(1+\beta t^{1/4})$ where $\beta$ is a constant with appropriate dimension of temperature. The heat capacity of metal is:

• A. $\frac{4P{\left(T\left(t\right)-T_0\right)}^3}{{\beta }^4{T_0}^4}$
• B. $\frac{4P{\left(T\left(t\right)-T_0\right)}^2}{{\beta }^4{T_0}^3}$
• C. $\frac{4P{\left(T\left(t\right)-T_0\right)}^3}{{\beta }^4{T_0}^5}$
• D. $\frac{4P{\left(T\left(t\right)-T_0\right)}^3}{{\beta }^4{T_0}^2}$

Answer 1

The correct option is A

$\frac{4P{\left(T\left(t\right)-T_0\right)}^3}{{\beta }^4{T_0}^4}$

Step 1. Given data:

Temperature as function of time,$T\left(t\right)=T_0(1+\beta t^{1/4})$ $\left(1\right)$

Step 2. Finding the heat capacity:

Heat capacity, $H=\frac{dQ}{dT}$

$H=\frac{dQ}{dT}\times \frac{dt}{dt}=P\times \frac{1}{{\displaystyle \frac{dT}{dt}}}$ $\left(2\right)$

Where $dQ$ is the heat energy supplied and $dT$ is corresponding change in temperature

$P$ is the the rate of heat energy

Differentiating equation $\left(1\right)$, we get

$\frac{dT}{dt}=T_0\left(\beta \frac{1}{4}{t}^{-3/4}\right)$ $\left(3\right)$

Also, rearranging equation $\left(1\right)$,

$T\left(t\right)-T_0=T_0\beta t^{1/4}$

$\Rightarrow t={\left(\frac{T\left(t\right)-T_0}{T_0\beta }\right)}^4$

Putting equation $\left(3\right)$ and value of $t$ in equation $\left(2\right)$, we get

$H=\frac{4P{\left(T\left(t\right)-T_0\right)}^3}{{\beta }^4{T_0}^4}$

Hence, option A is correct