# Heat Transfer in Freezer: Solve the Problem with John

• oldschool
In summary, to change from ice to a liquid, you need Q=(200g)(.5)(-5-0) (to change from ice to a liquid) + (200g)(79.7)(latent heat)+(200g)(1)(0-96)
oldschool
got a proplem that i just can't seem to get
200g of water is placed in a 80g aluminum cup and left undesturbed in a freezer until it reaches -5c the cup is then taken out and heated to 96c
a. how much heat is needed to increase the temp (from the -5c to 96c)
b. replace the alum cup with an iron cup of 120g and replace the water with ethyl alcohol how much alcohol is neded to add the same amount of heat as in part A and to raise the temp from -5c to 96c

thanks for any help
john

a) use the "specific heat" formula: Q = cm delta-T. But use it twice. FInd the total heat (Q_cup + Q_water) for the same change in T.

b) once you have the total Q, set up the same equation as before (except now for the specific heats of iron and alcohol). Use the total Q from part a and solve for the mass of alcohol.

so for a it would be Q=(200g)(.5)(-5-0) (to change from ice to a liquid) + (200g)(79.7)(latent heat)+(200g)(1)(0-96)

do i just add on the aluminum q then to that ?

Qalum=(80g)(.22)(-5-96)

looking at it i believe i do just add that on but i would like to make sure I am rite lol

Well, you don't simply have to increase the temperature. At 0°C, you have to melt the ice. That will require some amount of kJ/kg. Only have that heat is thrown into the ice, can you start to raise the temperature of the water some more.

So in addition to Qrequired to heat the water and Qrequired to heat the can, you also have Qrequired to melt the ice.

oldschool said:
so for a it would be Q=(200g)(.5)(-5-0) (to change from ice to a liquid) + (200g)(79.7)(latent heat)+(200g)(1)(0-96)

do i just add on the aluminum q then to that ?

Qalum=(80g)(.22)(-5-96)

looking at it i believe i do just add that on but i would like to make sure I am rite lol

then for b it would be (assuming the top is correct and i did cacls right it should be 37417.6 cals)

37417.6=(m)(.58)(-5-78)+(m)(25)+(m)(.58)(78-96)

that look correct?

oldschool said:
so for a it would be Q=(200g)(.5)(-5-0) (to change from ice to a liquid) + (200g)(79.7)(latent heat)+(200g)(1)(0-96)

do i just add on the aluminum q then to that ?

Qalum=(80g)(.22)(-5-96)

looking at it i believe i do just add that on but i would like to make sure I am rite lol

What you meant to say was

so for a it would be Q=(200g)(.5cal/g.oC)(0-(-5))oC + (200g)(79.7cal/g) (latent heat; to change from ice to a liquid))+(200g)(1cal/g.oC)(96-0)oC

and

Qalum=(80g)(.22cal/g.oC)(96-(-5))oC

oldschool said:
then for b it would be (assuming the top is correct and i did cacls right it should be 37417.6 cals)

37417.6=(m)(.58)(-5-78)+(m)(25)+(m)(.58)(78-96)

that look correct?

Here i assume you are talking about the alcohol, and I will assume you have the constants correct, except for the missing units (we all do that, but we shouldn't) and the temperature values being reversed again (this is what I wanted to call attention to more than the units). Of course you still have to find the heat for the cast iron cup. Plus, there might be a bit of a "trick" to this question. If the cast iron cup is open, how much heat is it going to take to raise the temperature of the cup above the vaporization temperature of the alcohol? Where is the alcohol vapor?

Last edited:
OlderDan said:
What you meant to say was

so for a it would be Q=(200g)(.5cal/g.oC)(0-(-5))oC + (200g)(79.7cal/g) (latent heat; to change from ice to a liquid))+(200g)(1cal/g.oC)(96-0)oC

and

Qalum=(80g)(.22cal/g.oC)(96-(-5))oC

Here i assume you are talking about the alcohol, and I will assume you have the constants correct, except for the missing units (we all do that, but we shouldn't) and the temperature values being reversed again (this is what I wanted to call attention to more than the units). Of course you still have to find the heat for the cast iron cup. Plus, there might be a bit of a "trick" to this question. If the cast iron cup is open, how much heat is it going to take to raise the temperature of the cup above the vaporization temperature of the alcohol? Where is the alcohol vapor?

thanks for catching me on the iron cup lol i forgot all about that i saw the proplem of revered temps when i did the equation so i fixed that the vaper point of alcohol is like 118 i think so thankfully i didn't ahve to worry about that and the cup wasn't supposed to loose any heat

thanks for the help though

NB:
yes I did forget about the latent heat of ice melting. But someone already caught it. Ah, well.

## 1. How does heat transfer occur in a freezer?

Heat transfer in a freezer occurs through a process called convection, where the cold air inside the freezer absorbs heat from the items stored in it. This heat is then released outside the freezer, causing the temperature inside to decrease.

## 2. Why is it important to solve the problem with John's freezer?

If the problem with John's freezer is not solved, it can lead to food spoilage and potential health risks. It can also cause an increase in energy consumption and higher electricity bills.

## 3. What are some possible causes of heat transfer issues in a freezer?

Possible causes of heat transfer issues in a freezer include a malfunctioning compressor, damaged door seals, or a buildup of frost or ice on the evaporator coils. These can all lead to a decrease in the freezer's efficiency and an increase in temperature.

## 4. How can the problem with John's freezer be solved?

The problem with John's freezer can be solved by checking and repairing any faulty components, ensuring proper insulation and sealing of the freezer, and regularly defrosting and cleaning the freezer to remove any buildup of frost or ice.

## 5. How can proper maintenance help prevent heat transfer issues in a freezer?

Proper maintenance, such as regularly cleaning and defrosting the freezer, checking and repairing any faulty components, and ensuring proper insulation and sealing, can help prevent heat transfer issues in a freezer. This can also prolong the lifespan of the freezer and save energy and money in the long run.

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