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A solid ball of mass 10 kg at 40°C is gently placed in a liquid of mass 20 kg at 20°C and of specific heat capacity 1 cal/gm-°C. When the thermal equilibrium is attained, temperature of the system is 35°C. The specific heat capacity of solid ball is. Neglect heat capacity of the vessel.

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2 $cal / gm /^{0} C$

$1$ $cal / gm /^{0} C$

3 $cal / gm /^{0} C$

6 $cal / gm /^{0} C$

answer is D.

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Detailed Solution

heat gain by liquid = heat loss by solid ball.

$\Rightarrow 20 \times 1 \times 15 = 10 \times S \times 5 \Rightarrow S = 6 cal / gm /^{0} C$

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