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(D) According to Newton's Law of Cooling,the rate of loss of heat $\left( \frac{dQ}{dt} \right)$ is directly proportional to the temperature differenc

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$20$

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(D) According to Newton's Law of Cooling,the rate of loss of heat $\left( \frac{dQ}{dt} ight)$ is directly proportional to the temperature difference $(\Delta 	heta)$ between the liquid and its surroundings,provided the temperature difference is small.Let $\left( \frac{dQ}{dt} ight)_1 = 60 \; cal/sec$ at $T_1 = 80^{\circ}C$.The room temperature $T_0 = 20^{\circ}C$.The temperature difference $\Delta 	heta_1 = T_1 - T_0 = 80 - 20 = 60^{\circ}C$.Let $\left( \frac{dQ}{dt} ight)_2$ be the rate of loss of heat at $T_2 = 40^{\circ}C$.The temperature difference $\Delta 	heta_2 = T_2 - T_0 = 40 - 20 = 20^{\circ}C$.Using the proportionality $\frac{(dQ/dt)_1}{(dQ/dt)_2} = \frac{\Delta 	heta_1}{\Delta 	heta_2}$:$\frac{60}{(dQ/dt)_2} = \frac{60}{20}$$\frac{60}{(dQ/dt)_2} = 3$$(dQ/dt)_2 = \frac{60}{3} = 20 \; cal/sec$.

$A$ liquid is filled in a vessel which is kept in a room with a temperature of $20^{\circ}C$. When the temperature of the liquid is $80^{\circ}C$,it loses heat at the rate of $60 \; cal/sec$. What will be the rate of loss of heat when the temperature of the liquid is $40^{\circ}C$ in $cal/sec$?

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