Solve Temperature Problem: 8.7 Deg C

[Kcoats]

The formation of condensation on a cold glass of water will cause it to warm up faster than it would have otherwise. If 4.5 gm of water condenses on a 122 gm glass containing 222 gm of water at 5.0 deg C, what will the final temperature be? Ignore the effect of the surroundings.

The answer for this is 14.9 deg C

But...I can't figure out how to get that answer. lol

Here is what I am doing, where do I go wrong?

(1 cal/g deg C)(222g)(Tf-5 degC) + (.2 cal/g deg C)(122g) (Tf-5 deg C)
(122cal/ deg C)(Tf-5 deg C) + (24.4 cal/deg C)(Tf-5 deg C)
(540 cal/g)=(146.4 cal/ deg C)(Tf-5 deg C)
/146.4 cal/ deg C
3.7=(Tf-5 deg C)
8.7=Tf

 

[Kcoats]

Trial and error. :)
I figured out how to work it.

Q=(4.5g)(540cal/g deg C)=2430 cal
(1 cal/g deg C)(222g)(Tf-5 degC) + (.2 cal/g deg C)(122g) (Tf-5 deg C)
(122cal/ deg C)(Tf-5 deg C) + (24.4 cal/deg C)(Tf-5 deg C)=246.4 cal/ deg c(tf-5 deg C)
2430 cal/246.4 cal/deg c=14.9 deg C

 

[thepatientmental]

It seems like you are on the right track, but there are a few mistakes in your calculations. Let's break it down step by step:

1. First, we need to calculate the total heat energy of the system before the condensation occurs. This can be done using the formula Q = mCΔT, where Q is the heat energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

For the glass, we have:
Q1 = (122g)(0.2 cal/g·°C)(5°C) = 122 cal

For the water, we have:
Q2 = (222g)(1 cal/g·°C)(5°C) = 1110 cal

The total heat energy before condensation is:
Qt = Q1 + Q2 = 122 cal + 1110 cal = 1232 cal

2. Next, we need to calculate the heat energy released during the condensation. This can be done using the formula Q = mL, where Q is the heat energy, m is the mass, and L is the latent heat of condensation.

For water, we have:
Q3 = (4.5g)(540 cal/g) = 2430 cal

3. Now, we can calculate the final temperature using the formula Qt = Q1 + Q2 + Q3:
1232 cal = (122g)(0.2 cal/g·°C)(Tf - 5°C) + (222g)(1 cal/g·°C)(Tf - 5°C) + (4.5g)(540 cal/g)
1232 cal = (24.4 cal/°C)(Tf - 5°C) + (222 cal/°C)(Tf - 5°C) + 2430 cal
1232 cal = (246.4 cal/°C)(Tf - 5°C) + 2430 cal
(1232 cal - 2430 cal) = (246.4 cal/°C)(Tf - 5°C)
-1198 cal = (246.4 cal/°C)(Tf - 5°C)
-1198 cal / 246.4 cal/°C = Tf - 5°C
-4.86°C = Tf - 5°C
Tf = 0.14°C

So, the final temperature