A heating element of mass 100 ,g and having s | vedclass

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(B) At steady state, the power absorbed by the element is equal to the power lost to the surroundings (Newton's law of cooling). Given, power absorbed

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$1.0$

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(B) At steady state, the power absorbed by the element is equal to the power lost to the surroundings (Newton's law of cooling). Given, power absorbed $P = 100 \,W$. Thus, the rate of heat loss at $127^{\circ}C$ is $100 \,J/s$. For a small temperature change from $127^{\circ}C$ to $126^{\circ}C$, we can assume the rate of heat loss remains approximately constant at $100 \,W$. Mass of element $m = 100 \,g = 0.1 \,kg$. Specific heat $s = 1 \,J/(g^{\circ}C) = 1000 \,J/(kg^{\circ}C)$. Heat released to cool from $127^{\circ}C$ to $126^{\circ}C$ is $Q = ms\Delta T = 0.1 \,kg 	imes 1000 \,J/(kg^{\circ}C) 	imes 1^{\circ}C = 100 \,J$. Time taken $t = Q/P = 100 \,J / 100 \,W = 1.0 \,s$.

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