Question

A current carrying wire heats a metal rod. The wire produces a constant power $P$ to the rod. The metal rod is enclosed in an insulated conatainer. It is observed that the temperature $\left(T\right)$ 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 the metal is:

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

Answer 1

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

As we know,
$\Delta Q=C\Delta T$
On differentiating the above equation w.r.t. time
$\frac{dQ}{dt}=C\frac{dT}{dt}$
$\frac{dQ}{dt}$ is the rate at which energy get dissipated.
$P=C\frac{dT}{dt}$-----(1)
As given,
$T\left(t\right)=T_0(1+\beta t^{1/4})$
$\frac{dT}{dt}=T_0\beta \times \frac{1}{4}\times t^{-\frac{3}{4}}$
$\frac{dT}{dt}=\frac{T_0\beta t^{-\frac{3}{4}}}{4}$
Putting this in equation (1), we get
$P=C\frac{T_0\beta t^{-\frac{3}{4}}}{4}$
$C=\frac{4P{t}^{3/4}}{T_0\beta }$----------(2)
From the given equation we have,
$t^{\frac{1}{4}}=\frac{T-T_0}{T_0\beta }$
$t^{\frac{3}{4}}={\left(\frac{T-T_0}{T_0\beta }\right)}^3$
Replacing in equation 2 we get
$C=\frac{4P{\left(\frac{T-T_0}{T_0\beta }\right)}^3}{T_0\beta }$
$C=\frac{4P(T-T_0{)}^3}{{\beta }^4{t}_0^4}$