Question

An object with a mass of `3 kg`, temperature of `210 ^oC`, and a specific heat of `24 J/(kg*K)` is dropped into a container with `36 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 `=0.1^@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=210-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=0.024kJkg^-1K^-1`

The mass of the object is `m_0=3kg`

The mass of the water is `m_w=36kg`

`3*0.024*(210-T)=36*4.186*T`

`210-T=(36*4.186)/(3*0.024)*T`

`210-T=2093T`

`2094T=210`

`T=210/2094=0.1^@C`

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