Question

An object with a mass of `5 kg`, temperature of `123 ^oC`, and a specific heat of `14 J/(kg*K)` is dropped into a container with `32 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 change in temperature is `0.05ºC`

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=123-T`

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

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

`C_o=0.014kJkg^-1K^-1`

`5*0.014*(123-T)=32*4.186*T`

`123-T=(32*4.186)/(5*0.014)*T`

`123-T=2511.6T`

`2512.6T=123`

`T=123/2512.6=0.05ºC`

As `T<100ºC`, the water does not evaporate.