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The mass, specific heat capacity and initial temperature of the sphere was 1000 gm, $\frac{1}{2}$ cal/gm°C and 80°C respectively. The mass of the liquid and the calorimeter are 900 gm and 200 gm, and initially both were at room temperature 20°C. Both calorimeter and the sphere are made of same material. If the steady-state temperature after mixing is found to be 40°C, then the specific heat capacity of unknown liquid, is (approximately)

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1 cal/gm°C

0.25 cal/gm°C

1.5 cal/gm°C

0.5 cal/gm°C

answer is C.

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Detailed Solution

$S_{2} = \frac{(1000) (\frac{1}{2}) (80^{0} - 40^{0})}{(900 + 200) (40^{0} - 20^{0})} = \frac{500 \times 2}{1100} c a l / g m^{0} c$

= 1 cal/gm°C

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