Mass of substance in a system (Heat transfer)

AI Thread Summary
A 2.00 kg cube of nitrogen at -220.0°C is added to a 1.00 kg mixture of ice and water at 0.00°C, resulting in a final temperature of -100.0°C. To solve for the amount of ice in the original mixture, the energy balance must be calculated, where heat gained equals heat lost. The initial calculations indicate that it requires 476,400 Joules to raise the nitrogen's temperature to -100°C. It is important to consider the phase changes of nitrogen, which is solid below -210°C, liquid between -210°C and -196°C, and gas above -196°C. The problem requires using three specific heats and two latent heats for accurate results.
Zoey Brown
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Homework Statement


A solid, 2.00 kg cube of nitrogen at -220.0oC is added to a 1.00 kg mixture of ice and

water at 0.00 oC. If the final temperature of the mixture is -100.0 oC, how much ice was

there in the original mixture?

Homework Equations


e=mct
e=mLv/f

The Attempt at a Solution


Don't know where to start
 
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Zoey Brown said:

Homework Statement


A solid, 2.00 kg cube of nitrogen at -220.0oC is added to a 1.00 kg mixture of ice and

water at 0.00 oC. If the final temperature of the mixture is -100.0 oC, how much ice was

there in the original mixture?

Homework Equations


e=mct
e=mLv/f

The Attempt at a Solution


Don't know where to start
heat gained = heat lost
 
[Mentor's Note: The quoted post was deleted, but I'm leaving the quote since it should help the OP.]

Fizzix_G said:
Hi Zoey,

First you have to start off by calculating how much energy it takes to make nitrogen go to -100 degrees Celsius. Try doing this and the number you should get is 476 400 Joules.
Did you remember that N2 is liquid below -196C and gas above that temperature?
Oops, worse than that:
solid below -210C, liquid -210 < T < -196, gas at T > -196.
Need 3 specific heats and 2 latent heats!
 
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