Water vapor Thermodynamics

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1. The problem statement, all variables and given/known data

How much water vapor (100 ° C) should be added to 120 g of ice (0 ° C) so that the final system has a final temperature of 35° C.


2. Relevant equations
Qvapor = lvapor * mvapor during condensation, where lvapor =2260 kJ/kg
Q1 = cwater*mvapor*ΔT during cooling, and cwater = 4,19 kJ/kg

Qmelting =lmelting * mice during melting, and lmelting =333 kJ/Kg
Q2 = cwater*mice*ΔT during heating


3. The attempt at a solution

Energy which is emitted by the vapor is given by: Qvapor + Q1 = 2260m + 4,19*m*100

Energy taken up/used by the ice is given by: Qmelting + Q2 =
333*0,120 + 4,19*0,120*35

I then set energy emitted = energy taken up by the ice and solved for m

2260m + 4,19m*100 = 333*0,120 + 4,19*0,120*35
m= 0,0215 kg or 21,5 g.

Do you think this is the correct way to solve this problem?
 

DrClaude

Mentor
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Your approach is correct, but your numbers are not:
Energy which is emitted by the vapor is given by: Qvapor + Q1 = 2260m + 4,19*m*100
 
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Ok. I don´t see which number could possibly be wrong? The temperature change is + 100 degrees, 4,19 kJ/kg is cwater, and 2260 is lvapor (according to my somewhat dated textbook).
 
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I tried calculating m using both -100 degrees and +135 degrees, and still got the wrong answer.
 

DrClaude

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I tried calculating m using both -100 degrees and +135 degrees, and still got the wrong answer.
Don't plug in numbers randomly.

What does ##\Delta T## stand for in the equation ##Q = c m \Delta T##?
 
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ΔT stands for Tfinal - Tinitial, but using Tfinal = 35 and Tinitial= 100 gives a ΔT = -65, and a m = 0,0289 g, which is also the wrong answer.
 

BvU

Science Advisor
Homework Helper
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Could it be you forgot the 4,19*0,120*35 ? Or the 4,19*m*65 ? My guess: the latter!
 

DrClaude

Mentor
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2,625
ΔT stands for Tfinal - Tinitial, but using Tfinal = 35 and Tinitial= 100 gives a ΔT = -65, and a m = 0,0289 g, which is also the wrong answer.
Careful with the sign. You have defined ##Q_1## as the amount of energy given by the condensed steam to the final mixture.
 
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I still don´t get it. Did you mean the negative sign? I tried using + 65, but the answer still turns out wrong (m = .00695). The answer is supposed to be 22,7 g.
 

DrClaude

Mentor
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Solving 2260m + 4,19m*65 = 333*0,120 + 4,19*0,120*35 for m, you should indeed get m=0.0227 kg.
 
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I must´ve pressed something wrong on my calculator. Was a bit hasty...as usual. Now I got the right answer though. Thanks!
 

BvU

Science Advisor
Homework Helper
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Fr0m the 28.9 g (the 0.0289 was kg) it follows you forgot the 4,19*m*65. Elementary, my dear Watson.
 

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