Question

An object with a mass of `2 kg`, temperature of `311 ^oC`, and a specific heat of `18 (KJ)/(kg*K)` is dropped into a container with `37 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 change in temperature is `=58.7^@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=311-T`

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

The specific heat of water is `C_w=4.186kJkg^-1K^-1`

The specific heat of the object is `C_o=18kJkg^-1K^-1`

The mass of the objest is `m_0=2kg`

The mass of the water is `m_w=37kg`

`2*18*(311-T)=37*4.186*T`

`311-T=(37*4.186)/(2*18)*T`

`311-T=4.30T`

`5.30T=311`

`T=311/5.30=58.7^@C`

As `T<100^@C`, the water does not evaporate