Question

A current carrying wire heats a metal rod. The wire provides a constant power $\left(P\right)$ 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 $T\left(t\right)=T_0(1+\beta t^{\frac{1}{4}})$, where $\beta$ is a constant with appropriate dimension while $T_0$ is a constant with dimension of temperature. The heat capacity of the metal is

• A. $\frac{4P{\left(T\right(t)-T_0)}^2}{{\beta }^4{T}_0^2}$
• B. $\frac{4P{\left(T\right(t)-T_a)}^4}{{\beta }^4{T}_0^5}$
• C. $\frac{4P{\left(T\right(t)-T_0)}^3}{{\beta }^4{T}_0^4}$
• D. $\frac{4P\left(T\right(t)-T_0)}{{\beta }^4{T}_0^2}$

Answer 1

The correct option is C $\frac{4P{\left(T\right(t)-T_0)}^3}{{\beta }^4{T}_0^4}$

Rate of heat transfer through metal rod is :
$\frac{dQ}{dt}=H\frac{dT}{dt}=P\left(\text{constant}\right).....\left(i\right)$
Also temperature variation is given as
$\begin{array}{c}T=T_0(1+\beta t^{1/4})\\ \therefore \frac{dT}{dt}=\frac{T_0\beta }{4}{t}^{-3/4}\end{array}.....\left(ii\right)$
By equation (i)
$H=\frac{P}{\left(\frac{dT}{dt}\right)}=\frac{4P}{\beta T_0}{t}^{3/4}$
Substituting the value of t from equation (ii), we get
$H=\frac{4P{(T-T_0)}^3}{{\left(\beta T_0\right)}^4}$