# An object with a mass of #1 kg#, temperature of #170 ^oC#, and a specific heat of #32 J/(kg*K)# is dropped into a container with #12 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

##### 1 Answer

#### 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/

12L * 1000

50136 J/oK * (T-0)^oK = 32J/⋅K * (170-T)oK

50136*T J = 5440 - 32*T ; 50168*T = 5440 ; deltaT = 9.22

So, the water increases from

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

12L * 1000

50136 J

Heat available from the object to 100oC: 1kg * 32J/kg⋅K * 70 oK = 2240J Therefore, not nearly enough to vaporize any of the water (that's why water is so good for calorimetry).