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(D) The energy absorbed by the water per unit time is given by $P_{in} = I \cdot A$, where $I = 700 \ W \ m^{-2}$ and $A = 0.05 \ m^2$.$P_{in} = 700 \

Vedclass · Accepted Answer

$8.33$

Vedclass · Answer

(D) The energy absorbed by the water per unit time is given by $P_{in} = I \cdot A$, where $I = 700 \ W \ m^{-2}$ and $A = 0.05 \ m^2$.$P_{in} = 700 	imes 0.05 = 35 \ W$.According to Newton's law of cooling, the rate of heat loss to the surroundings is given by $\frac{dQ}{dt} = k \cdot m \cdot s \cdot \Delta T$, where $\Delta T$ is the temperature difference between the water and the surroundings.After a long time, the system reaches a steady state where the rate of heat absorption equals the rate of heat loss:$P_{in} = \frac{dQ}{dt}$$35 = k \cdot m \cdot s \cdot \Delta T$Given $k = 0.001 \ s^{-1}$, $m = 1 \ kg$, and $s = 4200 \ J \ kg^{-1} \ K^{-1}$:$35 = 0.001 	imes 1 	imes 4200 	imes \Delta T$$35 = 4.2 	imes \Delta T$$\Delta T = \frac{35}{4.2} = \frac{350}{42} = \frac{25}{3} \approx 8.33 \ ^{\circ}C$.

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