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Heat Transfer Math Problem


tylerbrowner
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Lets say you had a cup that contains 1500 cc of water at 60 degrees Fahrenheit and you dropped 300 cc of ice that was 30 degrees Fahrenheit. if the water could not transfer heat to anything else other than the ice at what temp would the water and ice reach equilibrium. I was wondering if there was some kind of math equation that i could use to solve this.

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Lets say you had a cup that contains 1500 cc of water at 60 degrees Fahrenheit and you dropped 300 cc of ice that was 30 degrees Fahrenheit. if the water could not transfer heat to anything else other than the ice at what temp would the water and ice reach equilibrium. I was wondering if there was some kind of math equation that i could use to solve this.

 

Yes, there is. Though this sounds a lot like homework (other than the use of Fahrenheit) so I am hesitant to just give the answer.

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Convert the temperature to Celsius. Each unit of ice melting will lower the temperature of 80 units of water by 1 deg. Celsius. I'll assume the simplification that the ice is at 0 deg. Celsius. If below 0, you need to look up what is needed to raise it to 0 deg.

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A good idea would be to post your thoughts on what how to approach this question, rather than just post the question.

 

There is no need to convert to centigrade, the calculation can be done perfectly satisfactorily in fahrenheit.

 

Give us a clue what you already know.

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I failed in Engineering Thermodynamics thrice in a row. So i don't clearly get this question, this is either too simple or too complicated.

But i guess the solution would be Heat gained by ice = Heat lost by water. So at some point the mixture will reach equilibrium ( unless you mean something like ice being solid and still absorbing heat which isn't even possible as far as i know). The heat gained from converting ice from about -1 C or 30 F to Ice at 0C. Then converting Ice at 0 C to water at 0 C which is latent heat of ice. Then upto the equilibrium state. So the total heat gained by the ice is calculated and when it is equated with the heat lost by the hot water i guess the equilibrium state will be achieved? Heat lost by the ice can be calculated when it's mass is multiplied with the specific heat of ice which is 2000 J/(kg °C). And the latent heat of ice is 3 x 105 J/kg which can be used to calculate energy gained from 0 C to 0 C water by multiplying the mass of ice with latent heat of ice. Then the gained by 0C water to the equilibrium state.

So something like

(Mass of ice * specific heat of ice* temp difference which is -1 c to 0 c) +( Mass of ice * latent heat of ice) + (Mass of water* specific heat of water * temp difference which is Te Equilibrium temperature - 0 C) = (Mass of water * specific heat of water * temp difference which is about 15C - Te)

Like i said i failed in Thermodynamics thrice, not kidding. So correct me if i am wrong.

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I failed in Engineering Thermodynamics thrice in a row. So i don't clearly get this question, this is either too simple or too complicated.

But i guess the solution would be Heat gained by ice = Heat lost by water.

 

Bingo! Energy is conserved. That's the basic concept.

 

The rest looks OK as well. Melting the ice takes a different calculation for its energy change than changing the temperature of the water, and you also have to account for getting the ice to its melting point.

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