Latent of Vaporization and Fusion

Question

A.) Two liters of boiling water are mixed with ice and placed in the freezer section of a refrigerator. Ignoring the cooling action of the freezer, how many grams of ice can the water melt?

B.) A camperer's ice chest is filled with two 4.00-kilograms bags of ice. In ten hours the ice turns completely to liquid. How many calories of heat must enter the ice chest each minute on order to melt the ice at this rate?

Answer 1

Hello Ivan!

I'll try and help out with your questions here, don't hesitate to let me know if you have any more questions at the end though!

Question #1

Two liters of boiling water are mixed with ice and placed in the freezer section of a refrigerator. Ignoring the cooling action of the freezer, how many grams of ice can the water melt?

Since we know that the 2L of water are boiling, we know that their initial temperature is 100°C.

In this problem we are using the thermal energy contained within these 2L of hot water in order to melt solid ice into liquid water, and are asked what is the maximum mass of ice that can be melted with this thermal energy.

In thermodynamics there is an energetic cost associated with making a material change phase from a more condensed phase to a less condensed phase (ex. solid --> liquid). This energetic cost is called a latent heat. For ice, the latent heat of melting is 334 kJ / kg, meaning that it "costs" 334 kJ of energy to convert one kilogram of solid ice into liquid water when it is at its melting temperature. Now, for an arbitrary mass of ice, the total amount of energy required to melt it would be (334 kJ/kg) * m_ice.

Now, let's analyze the amount of thermal energy in the hot water...

If we imagine extracting all available thermal energy of of the 2L of boiling water, that would mean cooling that water all the way from its start temp of 100°C down to the melting/freezing temperature of the ice at 0°C. We can quantify that amount of thermal energy with the equation:

ΔU = c * m * ΔT

where ΔU is the change in energy, c is a property of the material called the specific heat (c_water = 4186 J/kgC), m is the mass, and ΔT is the change in temperature (ΔT = 0ºC - 100ºC = -100ºC). To calculate the mass of 2L of water, we need the density of water, which is conveniently equal to 1 kg / L, so m = 2 kg.

ΔU = c * m * ΔT = (4186 J/kgC) * (2kg) * (-100ºC) = -837,200 J = -837.2 kJ

So cooling the 2 kg of water down by 100 degrees would require removing 837,000 J from it. Or, put another way, there is 837,000 J of thermal energy in the boiling water that is available for being used to melt ice. Now let's equate the two terms...

(Amount of Available Thermal Energy Stored in Boiling Water) = (Amount of Energy Used to Melt Ice)

837.2 kJ = (334 kJ/kg) * m_ice Now divide both sides by 334 kJ/kg....

m_ice = 2.51 kg

So the most ice that you could melt would be 2.51 kg.

Question #2

A camper's ice chest is filled with two 4.00-kilograms bags of ice. In ten hours the ice turns completely to liquid. How many calories of heat must enter the ice chest each minute on order to melt the ice at this rate?

We are dealing with a similar concept here -- the amount of energy required to change ice from solid to liquid phase. This will require the latent heat of melting term from earlier (334 kJ/kg).

We have two 4 kg bags of ice, so a total of 8 kg of ice. The total energy to convert that to liquid phase is (334 kJ/kg) * (8 kg) = 2672 kJ.

We are told that the melting process happens gradually, and takes ten hours to occur, so we can calculate that the rate of heat flow into the cooler must be (2672 kJ) / (10 hours) = 267.2 kJ / hour

This is the quantity that they want for our answer -- but it is in units of kJ / hour instead of calories / minute. Let's convert.

1 kJ = 1000 J

4.184 J = 1 calorie

1 hour = 60 minutes

267.2 kJ / hour = 1064 calories / minute

So 1064 calories of heat must flow into the cooler every minute on average.

Hope that helps! Let me know if you have other questions!

--Zach