Question

An object with a mass of `1 kg`, temperature of `120 ^oC`, and a specific heat of `32 J/(kg*K)` is dropped into a container with `12 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 `=0.08^@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=120-T`

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

The specific heat of the object is `C_o=0.032kJkg^-1K^-1`

The mass of the object is `m_0=1kg`

The volume of water is `V=12L`

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

The mass of the water is `m_w=rhoV=12kg`

`1*0.032*(120-T)=12*4.186*T`

`120-T=(12*4.186)/(1*0.032)*T`

`120-T=1569.8T`

`1570.8T=120`

`T=120/1570.8=0.08^@C`

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