Question

An object with a mass of `5 kg`, temperature of `163 ^oC`, and a specific heat of `14 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 temperature of the water changec by `0.09º`C

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=163-T`

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

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

`C_o=14Jkg^-1K^-1`

`m_0 C_o*(163-T) = m_w* 4186 *T`

`5*14*(163-T)=32*4186*T`

`160-T=(32*4186)/(70)*T`

`163-T=1913.6T`

`1914.6T=163`

`T=163/1914.6=0.09ºC`

The water does not evaporate