An insulated beaker with negligible mass contains liquid water with a mass of 0.

Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.310 kg and a temperature of 91.2 degrees C.


A) How much ice at a temperature of -10.2 degrees C must be dropped into the water so that the final temperature of the system will be 37.7 degrees C?

Take the specific heat of liquid water to be 4190 J/kg * K, the specific heat of ice to be 2100 J/kg cdot K, and the heat of fusion for water to be 334 kJ/kg.

Explanation / Answer

Heat loss = Heat gained => m*10.2*2100 + m*334000 + m*4190*37.7 = 0.31*4190*(91.2 - 37.7) m = 0.135 Kg

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