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On a winter day the temperature drops from -5 degree C to -15 degree C overnight. If a pan sitting outside contains 0.40kg of ice, how much heat is removed from the ice for the temperature change? (specific heat (ice) = .5kcal/kg) Is it 02.0kcal?

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  1. M*C*delta T = 0.4*0.5*10 = 2.0 kCal

    You are right!

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