An object with a mass of #2 kg#, temperature of #315 ^oC#, and a specific heat of #12 (KJ)/(kg*K)# is dropped into a container with #37 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?
1 Answer
The water does not evaporate. The final temperature of the water is:
So the temperature change:
Explanation:
The total heat, if both remain in the same phase, is:
Initial heat (before mixing)
Where
Now we have to agree that:
- The heat capacity of water is:
- The density of water is:
So we have:
Final heat (after mixing)
- The final temperature of both the water and the object is common.
- Also, the total heat is equal.
Therefore:
Use equation to find final temperature:
Provided the pressure is atmospheric, the water did not evaporate, since its boiling point is
So the temperature change: