Latent Heat Problem? Not sure how to solve

AI Thread Summary
To solve the latent heat problem of melting ice on a car window, the correct mass of the ice must be calculated using its volume derived from the window's dimensions. The volume of ice is found by converting the thickness from centimeters to meters and then multiplying by the area, followed by using the density of ice to find the mass in grams. The heat required to melt the ice involves calculating both the heat needed to raise the temperature of the ice to 0 degrees Celsius and the latent heat of fusion. The initial calculations provided were based on an incorrect mass estimate, leading to a lower than expected heat requirement. Accurate mass calculation is crucial for determining the correct total heat needed to melt the ice.
PBryan833
Messages
6
Reaction score
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
 
Physics news on Phys.org
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...
 
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
 
thank you both for the help!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top