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?

1 Answer
Mar 22, 2016

Answer:

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.