Question

An object with a mass of `1 kg`, temperature of `160 ^oC`, and a specific heat of `9 J/(kg*K)` is dropped into a container with `27 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 `=0.013ºC`

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=160-T`

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

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

`C_o=0.009kJkg^-1K^-1`

`1*0.009*(160-T)=27*4.186*T`

`160-T=(27*4.186)/(1*0.009)*T`

`160-T=12558T`

`12559T=160`

`T=160/12559=0.013ºC`

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