Question

An object with a mass of `3 kg`, temperature of `244 ^oC`, and a specific heat of `37 (KJ)/(kg*K)` is dropped into a container with `16 L` of water at `0^oC`. Does the water evaporate? If not, by how much does the water's temperature change?

Answer 1

Some water will evaporate.

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=244-T`

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

The specific heat of the object is `C_o=37Jkg^-1K^-1`

The mass of the object is `m_0=3kg`

The volume of water is `V=16L`

The density of water is `rho=1kgL^-1`

The mass of the water is `m_w=rhoV=16kg`

`3*37*(244-T)=16*4.186*T`

`244-T=(16*4.186)/(3*37)*T`

`244-T=0.603T`

`1.603T=244`

`T=244/1.603=152.2^@C`

As the final temperature is `T>100^@C`, some water will evaporate.