Thermodynamic process: freezing

[Any Help]

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?

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

 

[DrClaude]

What is the question you have to answer?

 

[Any Help]

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

 

[DrClaude]

Looks ok. (Although I do not like the lack of units in the intermediate steps of the calculations )