Thermodynamic process: freezing

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Homework Help Overview

The discussion revolves around a thermodynamic problem involving a 0.350 kg sample undergoing freezing in a cooling apparatus. The problem includes calculating the sample's heat of fusion and its specific heat in the frozen phase, given specific heat values and temperature changes.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the heat transfer during the cooling process and seeks verification of their calculations regarding heat of fusion and specific heat. Some participants question the clarity of the problem statement and the inclusion of units in calculations.

Discussion Status

Participants are engaged in verifying the calculations presented by the original poster. There is acknowledgment of the calculations, though some concerns about the presentation of units have been raised. The discussion is ongoing with no explicit consensus reached.

Contextual Notes

There is a note regarding the importance of including units in calculations, which may affect clarity and understanding of the problem. The original poster's request for confirmation of their answers indicates a focus on accuracy and comprehension of the thermodynamic principles involved.

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Homework Statement



A 0.350 kg sample is placed in a cooling apparatus that removes energy as heat at a constant rate. The figure below gives the temperature T of the sample versus time t; the horizontal scale is set by Ts = 80.0 min. The sample freezes during the energy removal. The specific heat of the sample in its initial liquid phase is 3260J/kg·K. (a)What are the sample’s heat of fusion and (b) its specific heat in the frozen phase?

upload_2016-11-3_8-0-41.png

Homework Equations


Q=mL
Q=mcdelta T
P=Q/t

The Attempt at a Solution

:[/B]
constat rate: Q=c.m.delta T=3260x0.35x30=34230 J
P=q/t=34230/(40x60)=14.2625 W
a) q=ml (negative since itis a solidification)
but q=pt=14.2625x30x60=2567.2=mL====> L=2567.2/0.35=73350 J/Kg = 73.35 KJ/Kg
b) q=cxmxdeltaT but Q=pt=14.2625x20x60=17115J ====> c=17115/(mx delta T)= 17115/(0.35x20)=2445 J/Kg.K
I want only to check myanswerif correct
 
Last edited:
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What is the question you have to answer?
 
DrClaude said:
What is the question you have to answer?
What are (a) the sample’s heat of fusion and (b) its specific heat in the frozen phase?
Sorry, I thought I copied it with the question
 
Looks ok. (Although I do not like the lack of units in the intermediate steps of the calculations :smile:)
 

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