A liquid of mass 250 g is kept warm in a vess | vedclass

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(A) According to Newton's law of cooling,the rate of heat loss is proportional to the temperature difference: $\frac{dq}{dt} = -k(T - T_0)$.In the fir

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$8000 J kg^{-1} K^{-1}$

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(A) According to Newton's law of cooling,the rate of heat loss is proportional to the temperature difference: $\frac{dq}{dt} = -k(T - T_0)$.In the first case,the heater maintains a constant temperature,so the power supplied equals the rate of heat loss: $P = k(T - T_0)$.Given $P = 30 W$,$T = 57^{\circ} C$,and $T_0 = 27^{\circ} C$,we have: $30 = k(57 - 27) \Rightarrow 30 = 30k \Rightarrow k = 1 W/K$.When the heater is switched off,the rate of heat loss is $\frac{dq}{dt} = ms \frac{dT}{dt} = -k(T - T_0)$.Here,$m = 250 g = 0.25 kg$,$T_{avg} = \frac{47 + 46.9}{2} = 46.95^{\circ} C$,$T_0 = 27^{\circ} C$,$\Delta T = 47 - 46.9 = 0.1^{\circ} C$,and $\Delta t = 10 s$.Substituting these values: $0.25 	imes s 	imes \frac{0.1}{10} = 1 	imes (46.95 - 27)$.$0.25 	imes s 	imes 0.01 = 19.95$.$0.0025s = 19.95 \Rightarrow s = \frac{19.95}{0.0025} = 7980 J kg^{-1} K^{-1}$.Rounding to the nearest option,$s \approx 8000 J kg^{-1} K^{-1}$.

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