Question

An object with a mass of `2 kg`, temperature of `244 ^oC`, and a specific heat of `33 (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

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

The specific heat of the object is `C_o=33kJkg^-1K^-1`

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

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

`2*33*(244-T)=16*4.186*T`

`244-T=(16*4.186)/(2*33)*T`

`244-T=1.015T`

`2.015T=244`

`T=244/2.015=121.1^@C`

As `T>100^@C`, the water will evaporate