Question

An object with a mass of `24 kg`, temperature of `210 ^oC`, and a specific heat of `7 J/(kg*K)` is dropped into a container with `48 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 rise by `0.15ºC`

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=210-T`

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

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

`C_o=0.007kJkg^-1K^-1`

`24*0.007*(210-T)=48*4.186*T`

`210-T=(48*4.186)/(24*0.007)*T`

`210-T=1388.6T`

`1389.6T=210`

`T=210/1389.6=0.15ºC`

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