How much thermal energy is released when 500g of steam at 100° C condenses into water and cools to 50°C
Givens: m=0.5 kg
t = 50
c = 2.010
Qtotal = Q1 + Q2
Qtotal = QmcΔt + mLv
Qtotal = (0.5kg)(2.010)(50) + (0.5)(2.3*10^6)
Qtotal = 1.15 * 10^6
The book says that the answer is 1.3*10^6?
What did I do wrong here?
Lv=2.256E6 J/kg is one error, not 2.3
Secondly, c=4.17J/g or 4180J/kgC is another.
Qtot=.5*4180*50+.5*2.256E6J/kg
= 1.23E6 J
now if you use the erronous 2.3KJ/kg, it becomes 1.25E6 J
then you get into the "significant digits" discourse, which technically should be one (500g, 100C, 50 g), and that gives even a more lubricious answer. Ignore the answer in the book, it was computed by a poorly paid graduate student, or worse.
The person above is absolutely right, T is (-50). But since the latent heat is being released that makes that value negative too, so it becomes -1.15E6 . So when you add latent heat + mct, the answer is -1.3E6. Whenever you are solving these types of questions, always remember that you add up the heat being released irregardless of the negative or positive sign associated with it to make the question less confusing.
why is t is 50 though? Isn't the initial temperature is 100 and t final is 50 and therefore, we need to subtract t2-t1 = -50??
Well, it looks like you lost some thermal energy on your way to the answer. Maybe it got stuck in traffic or got distracted by a funny meme. Who knows? But don't worry, I'm here to help you find it!
Now, let's take a closer look at your calculations. You correctly used the equation Qtotal = Q1 + Q2, which is a great start. However, I spotted a tiny error in your second term.
You correctly identified that Q2 represents the energy released when the steam condenses into water, and you used the formula Q2 = mLv, where m is the mass of the water (0.5 kg) and Lv is the latent heat of vaporization for water (2.3*10^6 J/kg). But there's a little typo in your calculation - you multiplied (0.5)(2.3*10^6) and got 1.15 * 10^6 J.
The correct calculation should be (0.5)(2.3*10^6) = 1.15 * 10^6 J.
Now, let's add up the two terms again:
Qtotal = (0.5 kg)(2.010 J/(kg°C))(50°C) + (0.5 kg)(2.3*10^6 J/kg)
Qtotal = 50.5 J + 1.15 * 10^6 J
Qtotal = 1.15 * 10^6 J + 50.5 J
And this grand total is...1.15 * 10^6 J + 50.5 J! So you were actually right the first time.
Now go celebrate with a nice cup of tea, or a cold glass of water if you like irony. Cheers!
c = 4180 .... thats whats wrong
To determine what went wrong, let's go through the calculations step by step:
1. You correctly identified the given values: m = 0.5 kg, t = 50°C, and c = 2.010 (specific heat capacity).
2. The first term in the equation Qtotal = Q1 + Q2 represents the thermal energy released when the steam cools from 100°C to 50°C. To calculate this, you used the equation Q1 = mcΔt.
Q1 = (0.5 kg)(2.010 J/g°C)(50°C) = 50.25 J
So far, your calculation is correct.
3. The second term in the equation Qtotal = Q1 + Q2 represents the thermal energy released during condensation. To calculate this, you used the latent heat of vaporization Lv and the mass m. However, you made a small mistake in using the wrong value for Lv.
Q2 = (0.5 kg)(2.3 × 10^6 J/kg) = 1.15 × 10^6 J
Based on your calculation, the total thermal energy released would be:
Qtotal = Q1 + Q2 = 50.25 J + 1.15 × 10^6 J = 1.15 × 10^6 J
Upon reviewing your calculations, it seems that you have made an error in copying the given value for Lv, which should be 2.3 × 10^6 J/kg. The corrected calculation will yield the correct answer:
Q2 = (0.5 kg)(2.3 × 10^6 J/kg) = 1.15 × 10^6 J
Therefore, the correct total thermal energy released when 500g of steam at 100°C condenses into water and cools to 50°C is indeed 1.15 × 10^6 J, which matches your initial calculation.
The answer provided in the book, stating 1.3 × 10^6 J, seems to be incorrect based on the given information and calculations.