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(C) The rate of heat supply is constant, so $\frac{dQ}{dt} = P$ (where $P$ is a constant).By definition, $C = \frac{dQ}{dT}$, so $dQ = C \cdot dT = aT

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(C) The rate of heat supply is constant, so $\frac{dQ}{dt} = P$ (where $P$ is a constant).By definition, $C = \frac{dQ}{dT}$, so $dQ = C \cdot dT = aT^3 \cdot dT$.Substituting this into the rate equation: $\frac{dQ}{dt} = aT^3 \frac{dT}{dt} = P$.Rearranging gives $T^3 \cdot dT = \frac{P}{a} \cdot dt$.Integrating both sides: $\int T^3 \cdot dT = \int \frac{P}{a} \cdot dt$.This yields $\frac{T^4}{4} = \frac{P}{a} \cdot t + 	ext{constant}$.Assuming $T = 0$ at $t = 0$, the constant is $0$, so $T^4 = \frac{4P}{a} \cdot t$.Thus, $T = \left( \frac{4P}{a} ight)^{1/4} \cdot t^{1/4}$.Since $T \propto t^{1/4}$, the graph of $T$ versus $t$ will be a curve that starts at the origin and has a decreasing slope as $t$ increases (concave down), which corresponds to Graph $C$.

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