Specific Heat Capacities/Latent Heat

[lando45]

OK, my teacher set me this question in preparation for my exams, but I was ill when he taught this topic back in November, so I don't really know how to go about answering it. I've tried conducting some research on the web but it hasn't really gotten me any further.

Suppose that water at room temperature of 28°C is put into an ideal refrigerator that maintains an inside temperature of -3°C. The specific heats of water and ice are, respectively, 4187 J/(kg·C°) and 2090 J/(kg·C°); the latent heat of fusion for water is 3.34x105 J/kg.

a) What is the cost of making 96 lb (43.2 kg) of ice if electricity costs $0.10 per kilowatt-hour?

b) What percentage of this cost goes into chilling the water from 28°C to 0°C (without freezing it)?

c) What percentage of the total cost goes into freezing this chilled water but not cooling the ice that forms?

d) What percentage of the total cost goes into cooling the ice that forms to -3°C?

So could somebody point me in the right direction please? He's given me only 14 more hours to answer it!

 

[Hootenanny]

What are your initial thoughts?

 

[lando45]

OK, well one kilowatt hour is 3,600,000J of energy, right? And the specific heat capacity of a substance is defined as the amount of energy required to raise or lower the temperature of 1kg of the substance by one degree? Well there is 43.2kgs of water so I multiplied the SHC (4187) of water by 43.2 to get 1180878.4J. So this is the amount of energy required to cool the water in the fridge by 1 degree. So I figured that multiplying that by 28 would give me the amount of energy required to get the water down to freezing point (0 degrees), which would be 5064595.2J - if I then convert this into kilowatt hours by dividing it by 3,600,000 I get 1.406832. It then says that it costs $0.10 per kWh, which would make the cost $0.14, is this right?

 

[Hootenanny]

Almost, you forgot about the latent heat, once you have cooled the water to 0 degrees c you must then change it's stage into ice.

-Hoot

 

[lando45]

OK so if I had the 334,000J from the latent heat onto the initial 5064595.2J I get 5398595.2J - converting this into kWh gives me 1.499609778 - at $0.10 per kWh this would cost $0.1499, is this right?

 

[Hootenanny]

I get 1.44 \times 10^{7} J for the latent heat. Are you using;

Q = m\cdot L_{f}

 

[lando45]

Ah I made a mistake, I forgot to multiply the 334000 by 43.2. Doing so gives me 14,428,800 like you said. So now if I add this to the original I get 19,493,395.2 - then if I divide this by 3,600,00 i get 5.41, multiplt by 0.1 and I end up with $0.54 - you think this is correct? Or am i still doing something wrong?

 

[lightgrav]

looks right ... but you MUST keep better track of what UNITS you have ...

4187 J/kgK . . . 1180878 J/K . . . 334,000 J/kg . . . 19,493,395 J . . .
. . . 3,600,000 J/kWhr . . . 0.1 $/kWhr . . .

to avoid this kind of error.

 

[lando45]

OK I tried $0.54 and that's wrong. Anyone know why?

 

[lightgrav]

(5064595 + 14428800 + 270864) J = 19764259 J /3600000 J/kWhr = 5.49 kWhr
which costs $0.549 .
Is your ice still at 0 celsius?

 

[lando45]

OK, here are all my calculations so far:

43.2kg x 4187J x 28degrees = 5,064,595.2J
43.2kg x 3.34 x 10^5 = 14,428,800J
Total = 14,428,800J + 5,064,595.2J = 19,493,395.2J
Convert into kWh: 19,493,395.2 / 3,600,000 = 5.414832 kWh
At cost of $0.10 per kWh, total would be: 5.414832 x $0.10 = $0.54

But this is wrong! WHY?!

 

[lando45]

(5064595 + 14428800 + 270864) J = 19764259 J /3600000 J/kWhr = 5.49 kWhr
which costs $0.549 .
Is your ice still at 0 celsius?

I thought I didn't need to include the ice part? The question asks "What is the cost of making 96 lb (43.2 kg) of ice if electricity costs $0.10 per kilowatt-hour?" - so surely cooling the water down to 0 degrees then doing the latent thing will make the ice? I don't need the SHC of ice for part a)?

 

[lightgrav]

The author later asks about percentages of "this ... total" .
I would bet he's treating the entire process as making the ice.

 

[Hootenanny]

If the refigerator maintains a constant temp of -3, then that means any ice made will have to be cooled to this temperature, as lightgrav said. Not one of the best worded problems in the world