An object with a mass of 4kg, temperature of 140oC, and a specific heat of 15Jkg⋅K is dropped into a container with 24L of water at 0oC. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
David Drayer
Aug 29, 2016

by my calculations the water will rise .08 degrees C. It will not evaporated not even close.

Explanation:

The object dropped into the water is going release energy into the water that the water will absorb. The object will lose heat energy in joules and the water will gain heat energy in joules.

The object will lose 4 x 15 joules / kg = 60 joules per 1Ko

1Co = 1Ko

so 140-0 = 140 Ko

60 x 140 = 8400 joules of energy released by the object.

Water has a heat capacity of 4186 J/ kg Ko

so 84004186 = 2.007 Ko

This is the increase for 1 Kg of water.

l liter of water at O Co and 1 atmosphere of pressure has a mass of 1 Kg

so 24 liters of water has a mass of 24 Kg.

2.007 Co for 1 Kg / 24 Kg = .084 Co increase.

The water is still effectively freezing.

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