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(D) According to the principle of calorimetry,heat lost by the aluminium sphere equals the heat gained by the water and the calorimeter.Heat lost by a

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$434$

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(D) According to the principle of calorimetry,heat lost by the aluminium sphere equals the heat gained by the water and the calorimeter.Heat lost by aluminium: $Q_{lost} = m_{Al} \cdot S_{Al} \cdot \Delta T_{Al} = 0.20 \cdot S_{Al} \cdot (150 - 40) = 0.20 \cdot S_{Al} \cdot 110 = 22 \cdot S_{Al}$.Heat gained by water: $Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T_{water} = 0.150 \cdot 4200 \cdot (40 - 27) = 0.150 \cdot 4200 \cdot 13 = 8190\, J$.Heat gained by calorimeter: $Q_{cal} = W \cdot c_{water} \cdot \Delta T_{cal} = 0.025 \cdot 4200 \cdot (40 - 27) = 0.025 \cdot 4200 \cdot 13 = 1365\, J$.Equating the heat: $22 \cdot S_{Al} = 8190 + 1365 = 9555$.$S_{Al} = \frac{9555}{22} \approx 434.31\, J/kg\cdot ^\circ C$.Rounding to the nearest integer,the specific heat of aluminium is $434\, J/kg\cdot ^\circ C$.

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