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(B) The rate of cooling is given by $P = \frac{dQ}{dt} = mS \left| \frac{dT}{dt} \right|$,where $P$ is constant power,$m$ is mass,and $S$ is specific

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$3 : 6 : 2$

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(B) The rate of cooling is given by $P = \frac{dQ}{dt} = mS \left| \frac{dT}{dt} \right|$,where $P$ is constant power,$m$ is mass,and $S$ is specific heat capacity.Thus,$S = \frac{P}{m} \left| \frac{dt}{dT} \right|$. Since $P$ and $m$ are constant,$S \propto \left| \frac{dt}{dT} \right|$.For the gas phase (from $t=0$ to $t=10$ s,$T=100^{\circ}C$ to $70^{\circ}C$): $\Delta t = 10$ s,$\Delta T = 30^{\circ}C$. So,$S_{gas} \propto \frac{10}{30} = \frac{1}{3}$.For the liquid phase (from $t=20$ to $t=40$ s,$T=70^{\circ}C$ to $50^{\circ}C$): $\Delta t = 20$ s,$\Delta T = 20^{\circ}C$. So,$S_{liquid} \propto \frac{20}{20} = 1$.For the solid phase (from $t=80$ to $t=100$ s,$T=50^{\circ}C$ to $10^{\circ}C$): $\Delta t = 20$ s,$\Delta T = 40^{\circ}C$. So,$S_{solid} \propto \frac{20}{40} = \frac{1}{2}$.Now,$S_{solid} : S_{liquid} : S_{gas} = \frac{1}{2} : 1 : \frac{1}{3}$.Multiplying by $6$,we get $3 : 6 : 2$.

$A$ substance is cooled at a constant power. Its temperature vs time graph is shown. The value of $S_{solid} : S_{liquid} : S_{gas}$ is:

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