Calculating Water Vapor Requirements for Temperature Change of a System

[Attis]

Homework Statement

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.

Homework 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

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]

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

 

[Attis]

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).

 

[DrClaude]

The temperature change is + 100 degrees

That's the number that's incorrect.

 

[Attis]

I tried calculating m using both -100 degrees and +135 degrees, and still got the wrong answer.

 

[DrClaude]

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 $Q = c m \Delta T$?

 

[Attis]

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]

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

 

[DrClaude]

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.

 

[Attis]

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]

Solving 2260m + 4,19m*65 = 333*0,120 + 4,19*0,120*35 for m, you should indeed get m=0.0227 kg.

 

[Attis]

I must´ve pressed something wrong on my calculator. Was a bit hasty...as usual. Now I got the right answer though. Thanks!

 

[BvU]

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