Question

An object with a mass of `16 kg`, temperature of `180 ^oC`, and a specific heat of `5 J/(kg*K)` is dropped into a container with `48 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 the tempertaure is `=0.07^@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=180-T`

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

The specific heat of the object is `C_o=0.005kJkg^-1K^-1`

The mass of the object is `m_0=16kg`

The volume of water is `V=48L`

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

The mass of the water is `m_w=rhoV=48kg`

`16*0.005*(180-T)=48*4.186*T`

`180-T=(48*4.186)/(16*0.005)*T`

`180-T=2511.6T`

`2512.6=180`

`T=180/2512.6=0.07^@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.