An object with a mass of #4 kg#, temperature of #315 ^oC#, and a specific heat of #12 (KJ)/(kg*K)# is dropped into a container with #39 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Aug 9, 2017

No. The water increases from #0^oC to 70.7^oC.#


This is really a phase change calorimetry problem from chemistry.

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 * (12kJ)/(kg⋅K) * Delta T^oK = 48000 J * Delta T^oK#

#39L * 1000(cm^3)/L * 1.0g/(cm^3) = 39000g# water.
#39000g * 4.178 J/(g-^oK) = 162942 (J/K)# required to heat the water.

#162942 J xx (T_f - 0) = 48000J xx (315 - T_f)# solve for final temperature #T_f#

162942T J = 15120000 - 48000T ; 210942*T = 15120000

#T = 71.7^oC#

So, the water increases from #0^oC to 71.7^oC#, it does not evaporate.

CHECK: Calculate the amount of heat required to raise the water from 0oC to 100oC. That will tell you whether some of the water would evaporate or not, given the amount of heat available from the object.

Water specific heat is 4.178 J/g-oK. Water mass can be taken as 1.0g/cm^3 for this calculation.
39L * 1000cm^3/L * 1.0g/cm^3 = 39000g water. 39000g * 4.178 J/g-oK = 162942 J/oK
162942 J/oK * 100 oK = 16294200 J required to reach the water boiling point.
Heat available from the object to 100oC: 4kg * 12000J/kg⋅K * 315 oK = 15120000 J Therefore, not nearly enough to vaporize any of the water (that's why water is so good for calorimetry).