# 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

#### 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

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#**

116984*T J = 8400 - 24*T ; 117008*T = 8400 ; deltaT = 13.9

So, the water increases from