FINAL TEMPERATURE QUESTION
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FINAL TEMPERATURE QUESTION

[From: ] [author: ] [Date: 11-11-03] [Hit: ]
After a few minutes, the ice has melted and the temperature of the system has reached equilibrium .Specific heat of copper = 0.Specific heat of ice = 2.Specific heat of water (liquid) = 4.Delta Hºfus(H2O) = 6.......
This question is seriously starting to piss me off Ive tried it for 3 days now and can't answer it I've punched the walls until my knuckles bled thats how angry it makes me.

An 87.0g cube of ice at -10ºC is placed on a 12.0 kg block of copper at 25ºC, and the entire system is isolated from its surroundings. After a few minutes, the ice has melted and the temperature of the system has reached equilibrium . Calculate the final temperature using the following information:

Specific heat of copper = 0.385 J g-1 ºC-1
Specific heat of ice = 2.01 J g-1 ºC-1
Specific heat of water (liquid) = 4.194 J g-1 ºC-1
Delta Hºfus(H2O) = 6.01 kJ mol-1

The final answer is 17 degrees. I have tried ******* converting ice to water first and all that ******* bullshit and NOTHING ******* WORKS HOLY ****

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This is a three step problem. Keep in mind that the copper has a lot of stored heat, being 12.0 kg

First, warm up the ice. Find the energy needed to warm up the ice.
q=(87.0g)(2.01)(10) = 1748.7 J

Second, melt the ice. Find the energy needed to melt the ice (latent heat). Convert grams to moles
(87.0 g ice)(1 mole/18.02 g) = 4.83 moles
q = (4.83)(6.01) = 29.03 kJ, or 29,030 J

Lastly, the water is going to heat up to the unknown temp. All this time, the energy needed is being taken from the huge block of copper. Add up all the water energy and equal it to the copper energy.

1748.7 J + 29,030 J + (87.0)(4.194)(x-273) = (12,000)(0.382)(298-x)
1748.7 J + 29,030 J + 364.9x - 99611.7 = 1366032 - 4584x
4949x = 1434865
x = 290 K, or converting back to Celsius: 290 - 273 = 17ºC

Hope that helps!
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