Question

An object with a mass of `2 kg`, temperature of `214 ^oC`, and a specific heat of `12 (KJ)/(kg*K)` is dropped into a container with `26 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 `=36.7^@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=214-T`

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

The specific heat of the object is `C_o=12kJkg^-1K^-1`

The mass of the object is `m_0=2kg`

The volume of water is `V=26L`

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

The mass of the water is `m_w=rhoV=26kg`

`2*12*(214-T)=26*4.186*T`

`214-T=(26*4.186)/(2*12)*T`

`214-T=4.53T`

`5.53T=214`

`T=214/5.53=36.7^@C`

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