(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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(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 $Q = ms\Delta T = 0.1 \,kg \times 1000 \,J/(kg^{\circ}C) \times 1^{\circ}C = 100 \,J$. Time taken $t = Q/P = 100 \,J / 100 \,W = 1.0 \,s$.

Class 11 Physics 10-2.Heat Transfer Newton's Law of Cooling

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$A$ heating element of mass $100 \,g$ and having specific heat of $1 \,J/(g^{\circ}C)$ is exposed to surrounding air at $27^{\circ}C$. The element attains a steady state temperature of $127^{\circ}C$, while absorbing $100 \,W$ of electric power. If the power is switched off, then the approximate time taken by the element to cool down to $126^{\circ}C$ will be (neglect radiation). (in $\,s$)

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$0.1$
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10-2.Heat Transfer

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Newton's Law of Cooling

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In an experiment to verify Newton's law of cooling,a graph is plotted between the temperature difference $(\Delta T)$ of the water and surroundings and time as shown in the figure. The initial temperature of water is taken as $80^{\circ}C$. The value of $t_{2}$ as mentioned in the graph will be...........

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$A$ solid sphere and a hollow sphere of the same material and same radius are heated to the same temperature. They are placed in an environment of the same temperature. If the temperature difference between both and the surroundings is $T$,then .......

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