Question

An object with a mass of `16 kg`, temperature of `270 ^oC`, and a specific heat of `9 J/(kg*K)` is dropped into a container with `32 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 not evaporate and the change in temperature is `=0.29^@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=270-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=0.009kJkg^-1K^-1`

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

The mass of the water is `m_w=32kg`

`16*0.009*(270-T)=32*4.186*T`

`270-T=(32*4.186)/(16*0.009)*T`

`270-T=930.2T`

`931.2T=270`

`T=270/931.2=0.29^@C`

As the final temperature is `T<100^@C`, the water will not evaporate