Question

An object with a mass of `4 kg`, temperature of `240 ^oC`, and a specific heat of `15 J/(kg*K)` is dropped into a container with `24L` 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 `=0.14ºC`

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=240-T`

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

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

`C_o=0.015kJkg^-1K^-1`

`4*0.015*(240-T)=24*4.186*T`

`240-T=(24*4.186)/(4*0.015)*T`

`240-T=1674.4T`

`1675.4T=240`

`T=240/1675.4=0.14ºC`

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