Calculate Heat Removed from 130 g Steam to Ice

  • Thread starter Thread starter jagged06
  • Start date Start date
  • Tags Tags
    Heat Ice Steam
AI Thread Summary
To calculate the heat removed when 130 g of steam at 145°C is cooled and frozen into ice at 0°C, the specific heat of steam and water must be correctly applied. The first step involves calculating the heat lost as steam cools to 100°C using the specific heat of steam. Next, the heat released during condensation from steam to water at 100°C must be considered, followed by the heat lost as water cools to 0°C. Finally, the heat released during the phase change from water to ice at 0°C should be included. The error likely lies in not accounting for the latent heat of condensation and the phase change correctly.
jagged06
Messages
16
Reaction score
0

Homework Statement



How much heat must be removed when 130 g of steam at 145°C is cooled and frozen into 130 g of ice at 0°C. (Take the specific heat of steam to be 2.01 kJ/kg·K.)

Answer is in kcal.

Homework Equations



sht.gif


The Attempt at a Solution



I know steam turns back into water at 100°C

Q1=C1m(deltaT)
Q1=(2010 J//kg·K)(.13kg)(418.15K-373.15K)

In the second part, instead of using the specific heat of steam, I used the specific heat of water= 4.186 joule/gram °C

Q2=C2m(deltaT)
Q2=(4.186 J/g·°C)(130g)(100°C)

I then added Q1 to Q2 and came up with an answer in Joules.
1 joule = 0.000239005736 kilocalories

But my answer is wrong. Anyone see what I missed? Probably a conversion issue.
 
Physics news on Phys.org
Condensation (latent heat).
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

How much ice to cool down this water?
Replies
17
Views
5K
Heat Transfer. Ice and steam in a container (easy)
Replies
2
Views
4K
Laten heat of fusion calculation
Replies
13
Views
2K
1 kg of ice at -20°C is mixed with 1kg steam at 200°C......
Replies
9
Views
7K
Finding the Value of M in a Physics Homework Problem
Replies
12
Views
2K
Thermodynamics (final temperature, ice mixed with steam)
Replies
3
Views
9K
Equilibrium temperature of some ice and steam
Replies
9
Views
4K
Another HELP problem: Steam to Ice
Replies
5
Views
2K
Finding the final temperature of mixed ice and steam
Replies
10
Views
5K
Thermodynamics: Steam is added to an ice cube
Replies
16
Views
3K

Hot Threads

Recent Insights

Back
Top