Latent Heat Problem? Not sure how to solve

  • #1
6
0

Homework Statement



A window of a car is found to be covered with 0.8cm of ice. If the area of the window is 1 m2, and the ice is at -12 degrees Celsius, what is the minimum heat required to melt all the ice? Take the density of ice to be 900 kg/m3, the specific heat capacity of ice to be 0.5 cal/g*K, and the latent heat of fusion to be 80 cal/g. Answer in kcal.[/B]


Homework Equations



deltaQ=mcdeltaT
Latent Heat= deltaQ= Lm[/B]


The Attempt at a Solution


So first off, I am confused in terms of finding out the mass to use. I wasnt sure if it is .512 g by multiplying .8x.8.x8 for cm3, or if you are supposed to make 0.8cm into meters then get the volume of that and multiply it by 900kg/m3 and then make kg into grams. If it is the second method, then the mass would be .46 grams?

So then what i tried doing is add Q1 and Q2 to each other. Q1 for me was (0.512)(.5 cal/g K)(12) and then Q2 i have as (80 cal/g)(.512) Then i added these together to get 44.032 calories which would be 0.044032 kcal. This answer does not seem right however and I feel like it is too low. What am I doing wrong? Thank You So much
 

Answers and Replies

  • #2
NTW
302
26
First of all, please correctly compute the volume of the ice in cm3. I remind you that you have a surface of 1 m2 with a thickness of 0,8 cm. Once you have the volume, convert it to mass in grams by multiplying it by the density of the ice.

And once you have correctly calculated the mass, you can proceed with the heat that must be supplied...
 
  • #3
20,514
4,393
So first off, I am confused in terms of finding out the mass to use. I wasnt sure if it is .512 g by multiplying .8x.8.x8 for cm3, or if you are supposed to make 0.8cm into meters then get the volume of that and multiply it by 900kg/m3 and then make kg into grams. If it is the second method, then the mass would be .46 grams?
Method 2 is the correct approach, but your answer is wrong. Please show us your work.
So then what i tried doing is add Q1 and Q2 to each other. Q1 for me was (0.512)(.5 cal/g K)(12) and then Q2 i have as (80 cal/g)(.512) Then i added these together to get 44.032 calories which would be 0.044032 kcal. This answer does not seem right however and I feel like it is too low. What am I doing wrong? Thank You So much
Your approach here is correct, but, because your estimate of the mass is say low, so is this answer.

Chet
 
  • #4
6
0
thank you both for the help!
 

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