Question

An object with a mass of `1 kg`, temperature of `155 ^oC`, and a specific heat of `17 (KJ)/(kg*K)` is dropped into a container with `15 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 `=33ºC`

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=155-T`

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

`C_w=4.186kJkg^-1K^-1`

`C_o=17kJkg^-1K^-1`

`1*17*(155-T)=15*4.186*T`

`155-T=(15*4.186)/(1*17)*T`

`155-T=3.69T`

`4.69T=155`

`T=155/4.69=33ºC`

As `T<100ºC`, the water does not evaporate.