Calculating Heat Transfer and Latent Heat

  • Thread starter Thread starter sin_city_stunner
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating the total heat given off when 740 kg of water cools from 15 degrees to 0 degrees and freezes. The calculations show that the heat released during cooling is -46,620,000 J, while the heat released during freezing is 244,200,000 J, resulting in a total heat of 197,580,000 J. The confusion arises from the use of negative signs in the equations, with participants clarifying that the negative sign indicates energy being released during cooling and freezing. It is emphasized that regardless of whether water freezes or melts, the energy involved remains the same, but the signs differ based on the process. The discussion highlights the importance of understanding the physical implications of heat transfer and latent heat in thermodynamic calculations.
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

I try to solve a thermodynamics problem on heat transfer
Replies
3
Views
2K
Question about thermal physics -- Ice cubes melting in water
Replies
8
Views
3K
Calculate energy required to heat water using a steam coil
Replies
9
Views
2K
Latent heat and specific heat in insulated container
Replies
1
Views
2K
Heat question -- Ice cube in a cup of water....
Replies
7
Views
10K
Heat Transfer and Capacity: Understanding Specific Heat and Latent Heat
Replies
21
Views
4K
Quantity of heat, phase change of ice to water
Replies
7
Views
3K
Thermodynamics relating to temperature and phase changes
Replies
1
Views
1K
Thermodynamics adding ice to water problem with latent heat
Replies
1
Views
3K
Heat and temperature question with calorimetry
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top