Question

An object with a mass of `4 kg`, temperature of `341 ^oC`, and a specific heat of `12 (KJ)/(kg*K)` is dropped into a container with `42 L` of water at `0^oC`. Does the water evaporate? If not, by how much does the water's temperature change?

Answer 1

The water does not evaporate and the temperature changes by `73.1ºC`

Explanation:

The heat is transferred from the hot object to the cold water.

Let `T=` final temperature of the object and the water

For the cold water, `Delta T_w=T-0=T`

For the object `DeltaT_o=341-T`

`m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)`

`C_w=4.186kJkg^-1K^-1`

`C_o=12kJkg^-1K^-1`

`m_0 C_o*(341-T) = m_w* 4.186 *T`

`4*12*(341-T)=42*4.186*T`

`341-T=(42*4.186)/(48)*T`

`341-T=3.66T`

`4.66T=341`

`T=341/4.66=73.1ºC`

The water does not evaporate