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 :
$T\left(t\right)=T_0\left(1+\beta t^{1/4}\right)$
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. $\frac{4P{\left(T\left(t\right)-T_0\right)}^3}{{\beta }^4{T}_0^4}$
• B. $\frac{4P\left(T\left(t\right)-T_0\right)}{{\beta }^4{T}_0^2}$
• C. $\frac{4P{\left(T\left(t\right)-T_0\right)}^4}{{\beta }^4{T}_0^5}$
• D. $\frac{4P{\left(T\left(t\right)-T_0\right)}^2}{{\beta }^4{T}_0^3}$

Answer 1

Answer: (A)

Rate of heat transfer through metal rod is :
$\frac{dQ}{dt}=C\frac{dT}{dt}=P$ (constant) ...(1)
Also temperature variation is given as
$T=T_0\left(1+\beta t^{1/4}\right)$ ...(2)
$\therefore \frac{dT}{dt}=\frac{T_0\beta }{4}{t}^{-3/4}$
By equation (1)
$C=\frac{P}{\left(\frac{dT}{dt}\right)}=\frac{4P}{\beta T_0}{t}^{3/4}$
Substituting the value of t from equation (2), we get
$C=\frac{4P{\left(T-T_0\right)}^3}{{\left(\beta T_0\right)}^4}$