A 1.0 kg cube of ice is dropped into 1.0 kg of water, and when equilibrium is reached, there are 2.0 kg of ice at 0.0 degrees C. the initial temperature of the water was 0 degrees C, and was the original temperature of the ice?
Specific heat capacity of water = 4186 J/kg x C
Specific heat capacity of ice = 209 J/kg x C
Latent heat of fusion = 330000 J/kg
A) one or two degress below 0.0 C
B) -80 C
C) -160 C
D) -240 C
Any help on this would be great!
Specific heat capacity of water = 4186 J/kg x C
Specific heat capacity of ice = 209 J/kg x C
Latent heat of fusion = 330000 J/kg
A) one or two degress below 0.0 C
B) -80 C
C) -160 C
D) -240 C
Any help on this would be great!
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The ice was colder than the water, so the ice gained heat and the water lost it.
The water was at 0 deg and ended up as ice at 0 deg. So the only heat transfer from the water was the heat of fusion, 330000 J/kg * 1.0 kg.
The ice that was dropped rose in temperature by T degrees. That took T * 1.0 kg * 209 J/kg.
(Note: The value of 209 is not right. It's about a factor of 10 too small. Check the problem again).
These two amounts of heat are equal. The same heat the left the water entered the ice.
T * 1 * 209 = 330000 * 1
Solve for T.
The water was at 0 deg and ended up as ice at 0 deg. So the only heat transfer from the water was the heat of fusion, 330000 J/kg * 1.0 kg.
The ice that was dropped rose in temperature by T degrees. That took T * 1.0 kg * 209 J/kg.
(Note: The value of 209 is not right. It's about a factor of 10 too small. Check the problem again).
These two amounts of heat are equal. The same heat the left the water entered the ice.
T * 1 * 209 = 330000 * 1
Solve for T.