Question

An object with a mass of `18 kg`, temperature of `240 ^oC`, and a specific heat of `6 J/(kg*K)` is dropped into a container with `48 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 rise in temperature ot the water is `=0.13ºC`

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=240-T`

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

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

`C_o=6Jkg^-1K^-1`

`m_0 C_o*(240-T) = m_w* 4186 *T`

`18*6*(240-T)=48*4186*T`

`240-T=(48*4186)/(108)*T`

`240-T=1860.4T`

`1861.4T=240`

`T=240/1861.4=0.13ºC`

The water does not evaporate