# 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?

Mar 21, 2016

No. The water increases from ${0}^{o} C \to {9.22}^{o} C .$

#### 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/$g {-}^{o} K .$ Water mass can be taken as 1.0g/$c {m}^{3}$ for this calculation. Heat available from the object: 1kg * 32J/kg⋅K * delta${T}^{o} K$

12L * 1000$c {m}^{3}$/L * 1.0g/$c {m}^{3}$ = 12000g water. 12000g * 4.178 J/g-oK = 50136 J/oK required to heat the water.

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

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

So, the water increases from ${0}^{o} C \to {9.22}^{o} C .$, 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$/ {g}^{o} K$. Water mass can be taken as 1.0g/$c {m}^{3}$ for this calculation.

12L * 1000$c {m}^{3}$/L * 1.0g/$c {m}^{3}$ = 12000g water. 12000g * 4.178 J/g-oK = 50136 J${/}^{o} K$

50136 J${/}^{o} K$ * 100 oK = 5013600 J required to reach the water boiling point.

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).