Question

An object with a mass of `4 kg`, temperature of `350 ^oC`, and a specific heat of `6 J/(kg*K)` is dropped into a container with `28 L` of water at `0^oC`. Does the water evaporate? If not, by how much does the water's temperature change?

Answer 1

No. The water increases from `0^oC to 13.9^oC.`

Explanation:

To calculate an actual final water temperature we equate the two thermodynamic changes and solve for the common value of temperature. Water specific heat is 4.178 J/g-oK. Water mass can be taken as 1.0g/cm^3 for this calculation. Heat available from the object: 4kg * 6J/kg⋅K * delta `T^oK`

28L * 1000cm^3/L * 1.0g/cm^3 = 28000g water. 28000g * 4.178 J/g-oK = 116984 J/oK required to heat the water.

116984 J/oK * (T-0)`T^oK` = 24J/⋅K * (350-T)`T^oK`

116984T J = 8400 - 24T ; 117008*T = 8400 ; deltaT = 13.9
So, the water increases from `0^oC to 13.9^oC`, it does not evaporate.