Energy added to Ice- determine final temperature

  • Thread starter markdflip
  • Start date
  • #1
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Homework Statement



A 10 kg block of ice has a temperature of -8°C. The pressure is one atmosphere. The block absorbs 4.17 x10^6 J of heat. What is the final temperature of the liquid water?

Q=4.17x10^6J
T_0=-8
Mass=10kg

4184= specific heat of water
2000=specific heat of ice
333.5=latent heat of fusion

Homework Equations


Q=mc(deltaT)
Q=mL


The Attempt at a Solution





Q_1=mc(delta)T
Q_1=(10kg)(2000)(8°C)
Q_1=160,000J of energy was needed to raise the temperature to 0 degrees Celsius

Q_2=mL
Q_2=(10kg)(333.5)
Q_2=3335J of energy needed to change ice to water.

160,000J+3335J=163,335J of energy used.
4,170,000J-163,335J= 4,006,665J of energy left

4,006,665J=mc(delta)T
4,006,665J=(10kg)(4184)(delta)T
4,006,665/(10*4184)=deltaT
deltaT=95.761°C
95.761 degrees Celsius should be the final temperature according to my calculations.
I'm fairly sure the process is right, my money is there is something wrong with a coefficient, but I can't figure it out.
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
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check your units for the heat of fusion
 
  • #3
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oh, that was in J/g...so that would be 333,550 J/Kg
That makes more sense, 674,500J would be left for the third step opposed to 4,006,665J

674,500J/(10*4184)=deltaT
DeltaT=16.12 degrees Celsius.
Sound realistic?
 
  • #4
mgb_phys
Science Advisor
Homework Helper
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Yes - it takes an awful lot of energy to melt ice (hence the icecream headache)

quick tip - if you write a number down in a physics problem and it doesn't have a unit you are probably making a mistake!
 
  • #5
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Thanks for the help. I need to fix my habit of not writing labels with constants since I just get used to them cancelling out.

I replaced the specific energy wof ice with 2050 instead of 2000, and used the correct value of latent heat at 335000 and got 16.03, the correct answer.

Thank you so much for your help.
 

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