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?
No. The water increases from
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
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)
116984T J = 8400 - 24T ; 117008*T = 8400 ; deltaT = 13.9
So, the water increases from