Question

An object with a mass of `3 kg`, temperature of `270 ^oC`, and a specific heat of `18 J/(kg*K)` is dropped into a container with `36 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 increase in temperature of the water `=0.1ºC`

Explanation:

The heat transferred from the hot object, is equal to the heat absorbed by 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=270-T`

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

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

`C_o=18Jkg^-1K^-1`

`m_0 C_o*(270-T) = m_w* 4186 *T`

`3*18*(270-T)=36*4186*T`

`270-T=(36*4186)/(3*18)*T`

`270-T=2790.7T`

`2791.7T=270`

`T=270/2791.7=0.1ºC`