Removing heat energy from a block of ice

AI Thread Summary
To cool a block of ice measuring 4.50 cm x 4.30 cm x 6.20 cm from 0°C to -31.0°C, approximately 7285.5 joules of energy must be removed. The calculations involve determining the mass of the ice using its density and volume, followed by applying the heat energy formula Q = Mc(T_f - T_i). It's important to note the sign conventions in energy calculations, as a negative value indicates heat removal. For submission in a computerized homework system, the final answer should be presented as a positive value. Understanding these details ensures accurate representation of the energy change involved.
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Homework Statement



How much energy must be removed from a 4.50 cm x 4.30cm x 6.20cm block of ice to cool it from 0C to -31.0C?

The density of ice is 920 kg/m^3. You might need to refer to the textbook for other physical constants of ice.

Homework Equations



Q = Mc(T_f-T_i)

heat energy = mass*heat capacity * change in temp

M = Density*volume

The Attempt at a Solution



Q = Mc(T_f-T_i) = DVc(T_f-T_i)

Q = (920)(.045*.042*.062)(2180)(-31-0)= -7285.5 joules
 
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Is there a question you wish to ask?
 
Are you saying that this is right?
 
I am asking you if you wish to ask a question because your original posting does not include one. If you do, please ask and I (we) will do my (our) best to answer it.
 
Don't you need to know how much heat you need to pull out of it? Doesn't the method I have listed do that? It finds the heat energy removed doesn't it?
 
Yes, yes and yes. However, if this is an answer that you need to put in a computerized homework system, you need to understand the sign conventions.

If the question asks to find "the heat that goes in" and the number is negative, that means that heat actually goes out. In this case, the question is to find the heat that goes out, therefore a positive number is in order.
 
So its just 7285.5 joules then?
 
That would appear to be the case, but I am not the program that makes the decision whether you are right or wrong.
 
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