Heat energy transferred in converting ice into steam

Click For Summary

Homework Help Overview

The problem involves calculating the heat energy required to convert 0.10 kg of ice at -10 degrees Celsius into steam at 100 degrees Celsius, encompassing multiple phases of heating and phase changes.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the stages of heating ice to water and then to steam, questioning the calculations and the specific heat capacities used. There is an exploration of the latent heat values and their impact on the total energy calculation.

Discussion Status

Some participants have provided calculations for each stage of the process, leading to a total energy requirement that aligns with a reference value. There is acknowledgment of a potential miscalculation regarding the latent heat of vaporization, prompting further clarification and verification of values used.

Contextual Notes

Participants note the importance of checking calculations and the potential for misprints in reference materials, which may affect the understanding of the problem.

physiman
Messages
13
Reaction score
0
1. Homework Statement

Calculate the heat energy required to convert 0.10 kg of ice at -10 degrees celsius into steam at 100 degrees celsius.




2. Homework Equations

E = m c dt (1)

where

E = heat energy (kJ)

m = unit mass (kg)

c = specific heat capacity (kJ/kg oC)

dt = temperature change (Kelvin or degrees Celsius)





3. The Attempt at a Solution

Given in the question:

m = 0.1 kg
dt = 100-(-10) = 110 degrees celsius
c = 2100 J Kg^-1 K^-1 (specific heat capacity of ice)


So what I did was using the equation:

E = m c dt

E = (0.1)(2100)(110)

E = 23100 J

which would be 23.1 kJ



The problem is that the answer section of the book says 301.6 KJ which is way above my result.
What's the mistake I'm doing?

Help appreciated.

 
Physics news on Phys.org
Ice must be heated to 0C; ice must then be converted into liquid water at 0C;
Water at0C must be heated to 100C; water must be converted to vapor (steam) at 100C.
There are 4 stages, each requiring heat energy
 
Thanks for the hint.For Ice -> water:

m = 0.1kg c=2100 J kg^-1 K^-1 dt = 10 degrees (-10 -> 0)

therefore E= mc dt = 0.1(2100)(10) = 2100 J

Energy required to melt the ice:

E= specific latent heat of fusion of ice * mass
= 325000(0.1)
= 32500 J

and then for water to steam:

m = 0.1 kg c= 4200 J kg^-1 K^-1 (for water) dt= 100 degrees (0->100)

=> E= 0.1 (4200) (100) = 42000 J

Energy required to vaporize 0.1 kg of water:

E = specific latent heat of vaporization of water * mass
= 225000(0.1)
= 22500 J

then i add them

2100 + 32500 + 42000 + 225000

and get 301600 J which is 301.6 kJ.Thank you.
 
That looks good.
Well done
Do you see that by far the greatest amount of energy is in converting liquid water to steam?
 
SORRY... misprint, you have the latent heat of vaporisation to convert water to steam out by factor of 10
Latent heat of vaporisation = 2300000 J/kg
I have only just noticed... fortunately you made a mistake in your multiplication so your answer is correct6.
should always double check calculations !
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 7 ·
Replies
7
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
25
Views
2K