Question

An object with a mass of `1 kg`, temperature of `155 ^oC`, and a specific heat of `32 (KJ)/(kg*K)` is dropped into a container with `15 L` of water at `0^oC`. Does the water evaporate? If not, by how much does the water's temperature change?

Answer 1

At the end the water is at `52.32 °C`

Explanation:

The object temperature decreases from 155°C to the temperature T1 giving heat to the water
`Q= c_(po) M_0(T_0-T_1)= (32kJ)/(kgK)1kg (155-T_1)°C`.
Water takes this heat, warming from t= 0°C to the temperature T1.
`Q= c_(pw)M_w(T_1-0°C)=(4.186 kJ)/(kgK) 15kg ((T_1-0)°C`
(as one liter of water is one kg of water) equaling the lost heat with the gained heat, and resolving as a function of `T_1`, you find `T_1= 52.32 °C`

as `32(155-T_1)=4.186xx15xx(T_1-0)`

or `4.186xx15xxT_1+32T_1=32xx155`

or `(62.79+32)T_1=94.79T_1=4960`

or `T_1=52.32 °C`

Warning: There doesn't exist a solid object with a specific heat of `(32kJ)/(kgK)` . Only the ammonia among the liquid substances has specific heat bigger than water, which is one