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

1 Answer
Aug 29, 2016

Answer:

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 # 1 K^o#

# 1 C^o# = # 1 K^o#

so 140-0 = 140 #K^o#

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

Water has a heat capacity of 4186 J/ kg #K^o#

so # 8400/4186# = 2.007 #K^o#

This is the increase for 1 Kg of water.

l liter of water at O #C^o# and 1 atmosphere of pressure has a mass of 1 Kg

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

2.007 #C^o# for 1 Kg / 24 Kg = .084 #C^o# increase.

The water is still effectively freezing.