How much heat is released from water into ice.

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The discussion revolves around calculating the heat released when 250 g of water at 10 degrees Celsius freezes and cools to -10 degrees Celsius. The user initially miscalculated the heat transfer by not properly accounting for the different stages of the process: cooling to freezing, freezing, and cooling to the final temperature. After breaking down the calculations into three distinct stages and using the correct heat constants, the user arrived at a total heat release of approximately -2.177 x 10^5 J. The user expressed satisfaction after identifying and correcting their mistakes. The thread highlights the importance of careful stage-by-stage analysis in thermodynamic calculations.
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Homework Statement


How much heat is released when 250 g of water at 10 degrees celcius freezes into ice and cools to -10 degrees celcius?

Homework Equations


Q=mcΔt

The Attempt at a Solution



q= (250g)(4186)(283-273)=1.046*10^4 J

than q=(250g)(2090)(263-283)=-1.045*10^4 J

when i subtract the two i get 1*10^4

The answer is -9.86*10^4 what am i doing wrong?
 
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Three stages:
1. Cool to freezing point
2. Freeze
3. Cool to final temp

Each stage is governed by a different heat constant.
 
OK this is what I'm doing

1. cool to freezing point

q=(.250kg)(4186J/kgK)(273K-283K)=-1.046*10^4J

2.Freeze

MLf=(.250)(3.33*10^5)=8.32*10^4 J

3.Cool to final temp

Q= (.250kg)(2090j/kgK)(-283K-273K)=-2.90510*10^5

so when i sum up Q1+mLf+Q2= -2.177*10^5 J
Like seriously what am i doing wrong?
 
got it! i found the mistakes ! thanks! :D
 
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