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(D) According to Newton's law of cooling, the rate of heat loss is $\frac{dQ}{dt} = k(T - T_{surr})$.When the temperature is constant at $50^{\circ}C$

Vedclass · Accepted Answer

$1500$

Vedclass · Answer

(D) According to Newton's law of cooling, the rate of heat loss is $\frac{dQ}{dt} = k(T - T_{surr})$.When the temperature is constant at $50^{\circ}C$, the heat supplied equals the heat lost: $10 = k(50 - 20) = 30k$. Thus, $k = \frac{1}{3} \ W/^{\circ}C$.When the heater is switched off, the system cools. The average temperature during the cooling interval is $T_{avg} = \frac{35.1 + 34.9}{2} = 35^{\circ}C$.The rate of heat loss at this average temperature is $\frac{dQ}{dt} = k(T_{avg} - T_{surr}) = \frac{1}{3}(35 - 20) = \frac{15}{3} = 5 \ J/s$.In $1 \ minute$ $(60 \ seconds)$, the total heat lost is $Q = 5 \ J/s 	imes 60 \ s = 300 \ J$.The change in temperature is $\Delta T = 35.1^{\circ}C - 34.9^{\circ}C = 0.2^{\circ}C$.The heat capacity $C$ is given by $C = \frac{Q}{\Delta T} = \frac{300 \ J}{0.2^{\circ}C} = 1500 \ J/^{\circ}C$.

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