# 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?

Mar 22, 2016

No. The water increases from ${0}^{o} C \to {13.9}^{o} C .$

#### 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 ${T}^{o} K$

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)${T}^{o} K$ = 24J/⋅K * (350-T)${T}^{o} K$

116984T J = 8400 - 24T ; 117008*T = 8400 ; deltaT = 13.9
So, the water increases from ${0}^{o} C \to {13.9}^{o} C$, it does not evaporate.