Question 7 on Test 1: In order to make iced-tea, a 50.0 g ice cube at 0.0 degrees Celsius is added to 250mL of tea at 20.0 degrees Celsius. What is the final temperature of the iced-tea once it has reached thermal equilibrium? Assume no heat is transferred to or from the surroundings.
I solved it using: (16675.91565J)+(50g)(4.184J.(degrees Celsius)-1.g-1)(Tf)=-(250g)(4.184J.(degrees Celsius)-1.g-1)(Tf-20 degrees Celsius)
The answer I got was: Tf=3.4 degrees Celsius
Is this the correct setup and answer?
Correct Answer For Test 1 Question 7
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Re: Correct Answer For Test 1 Question 7
Yes, that is correct. The setup is qice=-qtea, which is nΔHfus + mCΔT = -mCΔT. You plugged in all the right numbers and got the correct final temperature.
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Re: Correct Answer For Test 1 Question 7
For the set up for this problem, how did you get 16675.91565 for the first part of the equation?
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Re: Correct Answer For Test 1 Question 7
Clarissa Molina 1D wrote:For the set up for this problem, how did you get 16675.91565 for the first part of the equation?
You're supposed to get the deltaH of fusion from the formula sheet and multiply it by the grams of ice, which you convert into moles:
(50g H2O/18.02 g/mol H20) * (6.01 kJ/mol) = 16.6759kJ or 16675.91565J.
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