Introduction

The relationship between the power (P) supplied to an object, its heat capacity (S), and the rate of change of its temperature ($dT/dt$) is given by the fundamental principle of calorimetry. The power is the rate at which heat energy ($Q$) is supplied, so $P = dQ/dt$. The heat capacity is defined as $S = dQ/dT$. Combining these, we get: $P = \frac{dQ}{dt} = \frac{dQ}{dT} \cdot \frac{dT}{dt} = S \frac{dT}{dt}$ From this, we can express the heat capacity as: $S = \frac{P}{dT/dt}$ Our goal is to find the heat capacity, S. Since P is given as a constant, we need to find the derivative of the temperature with respect to time, $dT/dt$. The options for S are expressed in terms of temperature T, so we will need to eliminate the time variable 't' from our final expression.

Step-by-step Solution

1. Differentiate the Temperature Function We are given the temperature of the metal rod as a function of time: $T(t) = T_0(1 + \beta t^{1/4})$ To find $dT/dt$, we differentiate this expression with respect to time 't': $\frac{dT}{dt} = \frac{d}{dt} \left[ T_0(1 + \beta t^{1/4}) \right]$ Since $T_0$ and $\beta$ are constants, we get: $\frac{dT}{dt} = T_0 \cdot \frac{d}{dt} [1 + \beta t^{1/4}] = T_0 \left[ 0 + \beta \frac{d}{dt}(t^{1/4}) \right]$ Using the power rule for differentiation ($\frac{d}{dx}x^n = nx^{n-1}$): $\frac{dT}{dt} = T_0 \beta \left( \frac{1}{4} t^{1/4 - 1} \right) = T_0 \beta \left( \frac{1}{4} t^{-3/4} \right)$ $\frac{dT}{dt} = \frac{T_0 \beta}{4} t^{-3/4}$

2. Express Time 't' in Terms of Temperature 'T' The expression for $dT/dt$ contains the variable 't', but the options for the heat capacity are in terms of T. We need to eliminate 't' from our expression. We can do this by rearranging the given temperature function: $T(t) = T_0(1 + \beta t^{1/4})$ Let's use T instead of T(t) for simplicity: $\frac{T}{T_0} = 1 + \beta t^{1/4}$ $\frac{T}{T_0} - 1 = \beta t^{1/4}$ $\frac{T - T_0}{T_0} = \beta t^{1/4}$ Isolating $t^{1/4}$: $t^{1/4} = \frac{T - T_0}{\beta T_0}$ Our expression for $dT/dt$ involves $t^{-3/4}$, which is $(t^{1/4})^{-3}$. So we take the reciprocal and cube the expression: $t^{-3/4} = \left( t^{1/4} \right)^{-3} = \left( \frac{T - T_0}{\beta T_0} \right)^{-3} = \left( \frac{\beta T_0}{T - T_0} \right)^3 = \frac{\beta^3 T_0^3}{(T - T_0)^3}$

3. Substitute to find dT/dt in terms of T Now, substitute this expression for $t^{-3/4}$ back into our equation for $dT/dt$: $\frac{dT}{dt} = \frac{T_0 \beta}{4} t^{-3/4} = \frac{T_0 \beta}{4} \left( \frac{\beta^3 T_0^3}{(T - T_0)^3} \right)$ $\frac{dT}{dt} = \frac{\beta^4 T_0^4}{4(T - T_0)^3}$

4. Calculate the Heat Capacity (S) Finally, we use the formula $S = P / (dT/dt)$: $S = \frac{P}{\frac{\beta^4 T_0^4}{4(T - T_0)^3}}$ $S = P \cdot \frac{4(T - T_0)^3}{\beta^4 T_0^4}$ Replacing T with T(t) to match the options' notation: $S = \frac{4P(T(t) - T_0)^3}{\beta^4 T_0^4}$

Conclusion

Comparing our result with the given options:

A: ${{4P{{(T(t) - {T_0})}^4}} \over {{\beta ^4}T_0^5}}$

B: ${{4P{{(T(t) - {T_0})}^3}} \over {{\beta ^4}T_0^4}}$

C: ${{4P(T(t) - {T_0})} \over {{\beta ^4}T_0^2}}$

D: ${{4P{{(T(t) - {T_0})}^2}} \over {{\beta ^4}T_0^3}}$

Our derived expression matches option B.