Question

An object with a mass of `6 kg`, temperature of `160 ^oC`, and a specific heat of `9 J/(kg*K)` is dropped into a container with `27 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, the final temperature is `=0.1ºC`

Explanation:

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=160-T`

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

`C_w=4186Jkg^-1K^-1`

`C_o=9Jkg^-1K^-1`

`m_0 C_o*(160-T) = m_w* 4186 *T`

`6*9*(160-T)=27*4186*T`

`160-T=(27*4186)/(54)*T`

`160-T=2093T`

`2094T=160`

`T=160/2094=0.1ºC`

The water does not evaporate