Question

An object with a mass of `4 kg`, temperature of `350 ^oC`, and a specific heat of `6 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 does not evaporate and the change in temperature is `=0.06^@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=350-T`

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

The specific heat of the object is `C_o=0.006Jkg^-1K^-1`

The mass of the object is `m_0=4kg`

The volume of water is `V=32L`

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

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

`4*0.006*(350-T)=32*4.186*T`

`350-T=(32*4.186)/(4*0.006)*T`

`350-T=5581.3T`

`5582.3T=350`

`T=350/5582.3=0.06^@C`

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