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(A) Given power $P = \frac{dQ}{dt}$.Heat capacity $S = \frac{dQ}{dT} = \frac{dQ/dt}{dT/dt} = \frac{P}{dT/dt}$.Given $T(t) = T_0(1 + \beta t^{1/4})$.Di

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

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(A) Given power $P = \frac{dQ}{dt}$.Heat capacity $S = \frac{dQ}{dT} = \frac{dQ/dt}{dT/dt} = \frac{P}{dT/dt}$.Given $T(t) = T_0(1 + \beta t^{1/4})$.Differentiating with respect to $t$:$\frac{dT}{dt} = T_0 \cdot \beta \cdot \frac{1}{4} t^{-3/4} = \frac{\beta T_0}{4} t^{-3/4}$.Substituting this into the expression for $S$:$S = \frac{P}{(\beta T_0 / 4) t^{-3/4}} = \frac{4P}{\beta T_0} t^{3/4}$.From the given equation,$\beta t^{1/4} = \frac{T(t) - T_0}{T_0}$.Therefore,$t^{1/4} = \frac{T(t) - T_0}{\beta T_0}$.Raising both sides to the power of $3$:$t^{3/4} = \left( \frac{T(t) - T_0}{\beta T_0} \right)^3$.Substituting $t^{3/4}$ back into the expression for $S$:$S = \frac{4P}{\beta T_0} \cdot \frac{(T(t) - T_0)^3}{\beta^3 T_0^3} = \frac{4P(T(t) - T_0)^3}{\beta^4 T_0^4}$.

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