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 ($T$) of the metal rod changes with time ($t$) as $T(t)=T_0(1+\beta t^{1/4})$ where $\beta$ is a constant with appropriate dimensions while $T_0$ is a constant with dimensions of temperature. The heat capacity of metal is :

Step 1: State the temperature relation and given conditions.

The temperature $T$ of the metal rod at time $t$ is given by:

$$T(t) = T_0(1 + \beta t^{1/4})$$

The constant power supplied to the insulated rod is $P$.

Step 2: Relate heat capacity to power.

The heat capacity $C$ is defined by the ratio of heat added to the temperature change.

$$C = \frac{dQ}{dT}$$

Since the power $P$ is constant and there is no heat loss, the heat supplied in time $dt$ is $dQ = P \, dt$.

$$C = \frac{P}{\frac{dT}{dt}}$$

Step 3: Calculate the derivative of temperature with respect to time.

Differentiate the temperature equation with respect to $t$.

$$\frac{dT}{dt} = \frac{T_0 \beta}{4} t^{-3/4}$$

Step 4: Express the heat capacity as a function of time.

Substitute the derivative into the expression for $C$.

$$C = \frac{4P}{T_0 \beta} t^{3/4}$$

Step 5: Express time in terms of temperature.

Rearrange the temperature equation to solve for $t^{1/4}$.

$$t^{1/4} = \frac{T - T_0}{T_0 \beta}$$

Cube both sides of the equation.

$$t^{3/4} = \frac{(T - T_0)^3}{\beta^3 T_0^3}$$

Step 6: Find the final expression for the heat capacity.

Substitute $t^{3/4}$ into the equation for $C$.

$$C = \frac{4P(T - T_0)^3}{\beta^4 T_0^4}$$

Answer: The heat capacity of the metal is $\dfrac{4P(T - T_0)^3}{\beta^4 T_0^4}$.