A solid ball of mass 3 kg at 10 °C is dropped in
3/2kg of water at 70 °C. The resulting temperature is 50 °C. This means that specific heat of solid ball is:-
(assume no heat loss in surroundings)
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Answer:
3/2×1×70-50
=3/2×20
=3×10=>30j /kg
Explanation:
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