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(B) The energy radiated from a black body is given by the Stefan-Boltzmann law: $U = \sigma A T^4 t$,where $\sigma$ is Stefan's constant,$A$ is the su

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The temperature of the water would increase from $10.0^{\circ} C$ to a final temperature of $18^{\circ} C$

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(B) The energy radiated from a black body is given by the Stefan-Boltzmann law: $U = \sigma A T^4 t$,where $\sigma$ is Stefan's constant,$A$ is the surface area,$T$ is the absolute temperature,and $t$ is the time.In the first case,the energy collected is $U_1 = \sigma A T^4 t$. The temperature rise is $\Delta T_1 = 11.0^{\circ} C - 10.0^{\circ} C = 1.0^{\circ} C$.In the second case,the new area is $A' = A/2$ and the new temperature is $T' = 2T$. The energy collected is $U_2 = \sigma (A/2) (2T)^4 t = \sigma (A/2) (16T^4) t = 8 \sigma A T^4 t = 8 U_1$.Since the energy absorbed by the water is proportional to the temperature rise $(\Delta Q = ms \Delta T)$,we have $\frac{\Delta T_2}{\Delta T_1} = \frac{U_2}{U_1} = 8$.Therefore,$\Delta T_2 = 8 	imes \Delta T_1 = 8 	imes 1.0^{\circ} C = 8.0^{\circ} C$.The final temperature of the water will be $10.0^{\circ} C + 8.0^{\circ} C = 18.0^{\circ} C$.

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