Question

An object with a mass of `6 kg`, temperature of `50 ^oC`, and a specific heat of `4 J/(kg*K)` is dropped into a container with `24 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 will not evaporate and the change in temperature is `=0.012^@C`

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=50-T`

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

The [
The specific heat of the object is `C_o=0.004kJkg^-1K^-1`

The mass of the object is `m_0=6kg`

The volume of water is `V=24L`

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

The mass of the water is `m_w=rhoV=24kg`

`6*0.004*(50-T)=24*4.186*T`

`50-T=(24*4.186)/(6*0.004)*T`

`50-T=4186T`

`4187T=50`

`T=50/4187=0.012^@C`

As the final temperature is `T<100^@C`, the water will not evaporate. We expect this result as the temperature of the object is `<100^@C` and the specific heat is low.