Calculating Final Temperature of Phase Change Using Heat Absorption

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Homework Help Overview

The discussion revolves around calculating the final temperature of a block of ice undergoing a phase change after absorbing a specific amount of heat. The subject area includes thermodynamics and heat transfer principles, particularly focusing on latent heat and specific heat capacities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between heat absorbed and temperature change, questioning the inclusion of latent heat in the calculations. There is an attempt to clarify the steps needed to account for both the heating of ice to its melting point and the phase change itself.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the problem by breaking down the heat absorption into distinct phases: heating the ice, melting it, and then heating the resulting water. There is recognition of the complexity involved in combining different forms of heat energy and the need for clarity in calculations.

Contextual Notes

Participants note the importance of understanding how to handle different units and the total energy available for the phase change, indicating that assumptions about the energy distribution are under consideration.

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1. A 10-kg block of ice has a temperature of -10°C. The pressure is one atmosphere. The block absorbs 4.12E6J of heat. What is the final temperature of the liquid water?



2. Q = mL and Q = cm/\T



3. I tried doing Q = cmT and plugging in the values above to get 88.4 degrees as the final temperature, and using 4186J/(kgCdegree) for c, but it's wrong. I am confused on these types of problems.

 
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you've ignored the latent heat required for the phase change from ice to liquid water...
 
I know, but I don't understand how it fits in.

mL + cmT = cmT?
 
you could try working out how much specific heat would be needed to raise it to melting point, then the latent heat needed to melt it, then subtract both these values from the total energy, and do one more Q=cmT calculation to work out the final temperature.
 
what is the total energy, and how would that work, subtracting Q-L if they both have different units?
 
sorry, bit ambiguous there. By total energy, I meant the number you were given for the total work applied.

I would approach this problem like this (there's probably a more elegant way, and if you want to leave the terms algebraic and cancel down etc, that might help. Given a calculator, many people don't find it necessary)

1. use Q=mcT to calculate how much heat one would have to apply to the water to heat it to 0 degrees (i.e T = 10). Call this Q_1 for example

2. use Q = mL to calculate the latent heat that will be required to change the material to a liquid. Call this amount of heat Q_2 perhaps.

3. now, Q_1 + Q_2 is the total heat you have "used" so far. Subtract this from the amount of heat that was applied in total: 4.12E6 - Q_1 - Q_2. Call the remaining amount of heat Q_3

4. finally, work out the temperature change you get from applying this remaining amount of heat Q_3 to the (now liquid) water, starting at 0 degrees.
 
thank you very much, you helped me a lot.
 

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