# 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: