What Will Be the Final Temperature of Water After Adding Steam?

  • Thread starter Thread starter Jm4872
  • Start date Start date
  • Tags Tags
    Reaction
Click For Summary

Discussion Overview

The discussion revolves around calculating the final temperature of water after adding steam to it in an insulated container. The scenario involves specific masses and temperatures of steam and water, and participants are attempting to solve a homework problem related to thermal energy transfer.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents an attempt to calculate the final temperature using the equations for heat transfer, but expresses confusion about their solution process.
  • Another participant questions the clarity of the initial problem statement, asking for clarification on the units and values provided.
  • A subsequent post reiterates the need for clarification, confirming the values as 5.00g of steam at 100 degrees Celsius and 100.0g of water at 27 degrees Celsius.
  • Another participant seeks clarification on the specific heat capacity value used in the calculations, questioning what "4.18" refers to.

Areas of Agreement / Disagreement

Participants do not appear to agree on the clarity of the problem statement or the calculations presented. Multiple questions and requests for clarification indicate that the discussion remains unresolved.

Contextual Notes

There are missing units in the initial problem statement, and participants express uncertainty about the definitions of the quantities involved, which may affect the understanding of the problem.

Jm4872
Messages
12
Reaction score
0

Homework Statement



What will be the final temperature of the water in an insulated container as the result of passing 5.00 of steam at 100.0 into 100.0 of water at 27.0?


Homework Equations



q=mct
q=nH

The Attempt at a Solution



I am completely lost as to how to solve for final temperature, I attempted it, but it didn't work, this is what I did.

n=5.00g/(18.02g/mol)=0.277 mol

q=(0.277 mol)*(40.6KJ/mol)=11.27KJ
11.27KJ=mct
11.27KJ=(100g)*(4.18)*(Tf-27)
(11.27/418)=Tf-27
Tf=(0.02696)+27=27.0

can someone please help, I am so lost :( I don't understand what it is that I'm doing wrong
 
Physics news on Phys.org
Jm4872 said:
passing 5.00 of steam at 100.0 into 100.0 of water at 27.0?

5 of what, 100 of what, 100 of what, 27 of what?
 
Borek said:
5 of what, 100 of what, 100 of what, 27 of what?

oops, I didn't realize that the values didn't show up.

it should be 5.00g, 100degrees celsius, 100.0g and 27 degrees celsius
 
Jm4872 said:
11.27KJ=(100g)*(4.18)*(Tf-27)

4.18 of what?
 

Similar threads

Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
26K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
6
Views
15K