Calculating Heat Transfer and Latent Heat

  • Thread starter Thread starter sin_city_stunner
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around a problem related to heat transfer and latent heat, specifically involving the cooling and freezing of water. The original poster attempts to calculate the total heat given off by 740 kg of water as it cools from 15 degrees Celsius to 0 degrees Celsius and subsequently freezes.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculations involving specific heat capacity and latent heat, questioning the use of negative signs in the equations. There is discussion about the implications of freezing versus melting in terms of energy transfer.

Discussion Status

The conversation is ongoing, with participants providing insights into the reasoning behind the signs used in the equations. Some participants express confusion about the treatment of energy in freezing and melting processes, while others attempt to clarify these concepts.

Contextual Notes

There is mention of a discrepancy between the original poster's calculations and the expected answer from a textbook, as well as a reference to the original poster's educational background regarding the treatment of negative signs in heat transfer equations.

sin_city_stunner
Messages
8
Reaction score
0
Hey all,
i got a problem I am stuck with and it'll probably be easy for you but i can't figure it out for some reason... It has to do with heat transfer and latent heat...

A tank containing 740 kg of water cools from 15 degrees to 0 degrees and completely freezes. What is the total heat given off? How many hours would a 1.2 kW eletric heater have to work to provide the equivilant heat the water gives off? (The heater provides 1.2 x 10^3 J/s of heat)

thanks

heres my work:

for part 1: Q=mcT
Q= (740kg)(4200 J/ kg C)(0 degrees- 15 degrees) 9specific heat capacity for water= 4200 J/kg C
= -46 620 000 J

Q= LfM
Q= (3.3x10^5 J/kg)(740 kg)
Q= 244 200 000 J

Total heat= -46 620 000J + 244 200 000J
=197 580 000

book says answer is 2.9 x 10^8 (with S.d.) I know how to do the second part but its this first part messing me up...
 
Last edited:
Physics news on Phys.org
What's the minus sign for? Why doesn't the answer from the second equation have a minus?
 
the minus sign is there because deltaT is equal to : final temperature-initial temperature...which in the first part is 0-15...there is no negative sign in the second part because the delta T is positive
 
Last edited:
In the second part (referring to Q = LfM), if the water didn't freeze, but melt, how would the equation show it (where would the plus/minus sign come from)?
 
its the same regardless because the energy will be the same whether you are freezing or melting the object...
 
Umm.. No. You're now saying that freezing and melting both bind energy (as in opposite to give off).
There's a difference between you heating water from 0 to 15 degrees and cooling from 15 to 0.

Thus, there must be a minus sign in the equation Q = LfM, when something freezes (logically it gives off energy).
 
really? that must be the problem then...my teacher never taught us about a negative sign...he just said use this equation and plug in the numbers... sigh...thanks for your help BTW
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
10K
Replies
21
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K