# An object with a mass of 4 kg, temperature of 140 ^oC, and a specific heat of 15 J/(kg*K) is dropped into a container with 24L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Aug 29, 2016

by my calculations the water will rise .08 degrees C. It will not evaporated not even close.

#### Explanation:

The object dropped into the water is going release energy into the water that the water will absorb. The object will lose heat energy in joules and the water will gain heat energy in joules.

The object will lose 4 x 15 joules / kg = 60 joules per $1 {K}^{o}$

$1 {C}^{o}$ = $1 {K}^{o}$

so 140-0 = 140 ${K}^{o}$

60 x 140 = 8400 joules of energy released by the object.

Water has a heat capacity of 4186 J/ kg ${K}^{o}$

so $\frac{8400}{4186}$ = 2.007 ${K}^{o}$

This is the increase for 1 Kg of water.

l liter of water at O ${C}^{o}$ and 1 atmosphere of pressure has a mass of 1 Kg

so 24 liters of water has a mass of 24 Kg.

2.007 ${C}^{o}$ for 1 Kg / 24 Kg = .084 ${C}^{o}$ increase.

The water is still effectively freezing.