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 $T (t) = T_{0}$ $(1 + β t^{1 / 4})$ where $β$ is a constant with appropriate dimensions while $T_{0}$ is a constant with dimensions of temperature. The heat capacity of metal is

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$\frac{4 P {(T (t) - T_{0})}^{3}}{β^{4} T_{0}^{4}}$

$\frac{4 P (T (t) - T_{0})}{β^{4} T_{0}^{2}}$

$\frac{4 P {(T (t) - T_{0})}^{2}}{β^{4} T_{0}^{3}}$

$\frac{4 P {(T (t) - T_{0})}^{4}}{β^{4} T_{0}^{5}}$

answer is A.

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Detailed Solution

$\begin{array}{l} P = m s \frac{d T}{d t} = (H) \frac{d}{d t} [T_{0} (1 + β t^{1 / 4})] = \frac{H T_{0} β}{4} t^{- 3 / 4} \dots . . (1) \\ here H is heat capacity \\ Given : T = T_{0} [1 + β t^{1 / 4}] \\ \Rightarrow \frac{T}{T_{0}} - 1 = β t^{1 / 4} \\ \Rightarrow t^{- 1 / 4} = \frac{β T_{0}}{T - T_{0}} \\ \Rightarrow t^{- 3 / 4} = {(\frac{β T_{0}}{T - T_{0}})}^{3} \dots \dots . . (2) \\ Substitute (2) in (1) \\ P = \frac{H T_{0} β}{4} \frac{β^{3} T_{0}^{3}}{{(T - T_{0})}^{3}} \\ \Rightarrow H = \frac{4 P {(T - T_{0})}^{3}}{β^{4} T_{0}^{4}} \end{array}$