Question

An object with a mass of `4 kg`, temperature of `281 ^oC`, and a specific heat of `12 (KJ)/(kg*K)` is dropped into a container with `25 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 `=88.4^@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=281-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=12kJkg^-1K^-1`

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

The mass of the water is `m_w=25kg`

`4*12*(281-T)=25*4.186*T`

`281-T=(25*4.186)/(4*12)*T`

`281-T=2.18T`

`3.18T=281`

`T=281/3.18=88.4^@C`

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