Heat and Phase change: latent heat

Click For Summary

Homework Help Overview

The problem involves calculating the heat required to melt ice on a car windshield, given specific conditions such as temperature, thickness, area, and density of ice. The subject area pertains to thermodynamics, specifically latent heat and phase changes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to consider two steps: raising the temperature of the ice to its melting point and then melting the ice. There are questions about the correct temperatures and methods to calculate heat transfer, including the use of specific heat and latent heat.

Discussion Status

Participants are exploring different interpretations of the problem, with some suggesting the use of specific heat for the initial temperature change and latent heat for the melting process. There is a mix of understanding regarding the roles of area and density in the calculations.

Contextual Notes

Some participants express uncertainty about the concepts, indicating that they are new to the material. There is a lack of consensus on the correct approach, and assumptions about temperature and heat transfer methods are being questioned.

scw287
Messages
51
Reaction score
0

Homework Statement



A woman finds the front windshield of her car covered with ice at -12.8°C. The ice has a thickness of 4.60 10-4 m, and the windshield has an area of 1.25 m2. The density of ice is 917 kg/m3. How much heat is required to melt the ice?

Homework Equations


Q=mL or change temp=Q/constant*mass

The Attempt at a Solution


Volume = Area * thickness<<then found the density
Q=mct Energy = mass* specific heat* cahnge in temp
heres the numbers that I plugged in: Q=(917)*(5.75*10^-4)*(12.8)*(22.6*10^5)<this last number is the latent fusion for water

the answer I got was wrong. not sure if I'm make a simple mistake or completely on the wrong track!

thanks in advance:wink:
 
Physics news on Phys.org
Try this. Think in terms of two separate steps:
(1) You must raise the temp of the ice to the melting point
(2) You must melt the ice

Find the heat needed for each step and add them up.
 
1.if i raise the temp of the ice to the melting pt. it would need to be at its boiling pt 100 degrees celsius. heat rises

2. now must melt the ice at 0 degrees celsius

I'm sorry I really don't know what I'm doing just learned this for the first time today along with four other physics chapter.

could I use m*L<heat gained
and c*m*change in temp.< heat lost
and set the two equal?
?
 
scw287 said:
1.if i raise the temp of the ice to the melting pt. it would need to be at its boiling pt 100 degrees celsius. heat rises
What temperature must the ice be in order to melt? Just melt, not boil!
 
well the melting point is at 0. degrees celsius
 
is the latent fusion the same thing as the temperature for which ice needs to be in order to melt?
 
scw287 said:
well the melting point is at 0. degrees celsius
Right. So how much heat is needed to raise the temperature of the ice? (Hint: Specific heat)

scw287 said:
is the latent fusion the same thing as the temperature for which ice needs to be in order to melt?
The latent heat of fusion tells you how much energy you need to melt something when it's already at the melting point.
 
Specific heat for water=4186
 
So I need to use the formula Q=mass*specific heat*change in temp for the first part right?
 
  • #10
scw287 said:
Specific heat for water=4186
What's the specific heat of ice?
scw287 said:
So I need to use the formula Q=mass*specific heat*change in temp for the first part right?
Right.
 
  • #11
good after I find that I need add that to melting the ice. How does the area and the density come into play?
 
  • #12
scw287 said:
How does the area and the density come into play?
It will allow you to find out how much ice you need to melt.
 
  • #13
I'm still not sure how to find the melting of the ice, do i need to use latent heat constant?
ahhh so add the two values together and then multiply by the area*the density?
 

Similar threads

Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K