Question

An object with a mass of `4 kg`, temperature of `360 ^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 final temperature is `=0.06ºC`

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=360-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*(360-T) = m_w* 4186 *T`

`4*6*(360-T)=32*4186*T`

`360-T=(32*4186)/(24)*T`

`360-T=5581.3T`

`5582.3T=360`

`T=360/5582.3=0.06ºC`

The water does not evaporate