How many calories of heat are released as 200.0g of water cool from a liquid at 25.0C to ice at -15.0C?

Md Ariful Islam

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Md Ariful Islam

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According to the heat equations: Q = C1mDelta;T1 + C2mDelta;T2 + lambda;m. m = 0.2 kg For water: C1 = 4.183x103 [J / (kg K)] Delta;T1 = 25 - 0 = 25 For ice: C2 = 2.11x103 [J / (kg K)] Delta;T2 = 0 - (-15) = 15 lambda; = 3.3x105 J /kg Thus Q = 0.2 (4.183x103*25 +3.3x105 + 2.11x103 *15) = 93.245 J In calories: Q = 22.29 kcal.
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