Question

An object with a mass of `3 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 will not evaporate and the change in temperature is `=0.04^@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=123-T`

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

The specific heat of the object is `C_o=0.014kJkg^-1K^-1`

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

The volume of water is `V=32L`

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

The mass of the water is `m_w=rhoV=32kg`

`3*0.014*(123-T)=32*4.186*T`

`123-T=(32*4.186)/(3*0.014)*T`

`123-T=3189.3T`

`3190.3T=123`

`T=123/3190.3=0.04^@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 not very high and the specific heat is low.