Water and turpentine oil (specific heat less | vedclass

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(C) The rate of cooling is given by $\frac{dT}{dt} = \frac{kA(T - T_0)}{mc}$,where $m$ is the mass and $c$ is the specific heat capacity.Since the mas

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$B$ and $A$ will represent the cooling curves of water and oil,respectively.

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(C) The rate of cooling is given by $\frac{dT}{dt} = \frac{kA(T - T_0)}{mc}$,where $m$ is the mass and $c$ is the specific heat capacity.Since the mass $m$ and the surface area $A$ are the same for both,the rate of cooling is inversely proportional to the specific heat capacity $c$,i.e.,$	ext{Rate of cooling} \propto \frac{1}{c}$.Given that $c_{	ext{oil}}  (	ext{Rate of cooling})_{	ext{water}}$.This means that the oil cools down faster than the water.Therefore,at any given time $t > 0$,the temperature of the oil will be lower than the temperature of the water.Looking at the graph,curve $B$ shows a faster drop in temperature compared to curve $A$.Thus,curve $B$ represents the cooling curve of oil,and curve $A$ represents the cooling curve of water.

Water and turpentine oil (specific heat less than that of water) are both heated to the same temperature. Equal amounts of these,placed in identical calorimeters,are then left to cool in the air. Which of the following statements is correct regarding their cooling curves?

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