I do not understand how the final temperature of the nail was so much higher than the final temperature of the water. It sounds like thermal equilibrium was not attained.
What exactly did the experiment consist of?
A reasonable entropy experiment might have consisted of quenching a hot nail in water, until the nail attained the temperature of the water. The heat lost by the nail would equal the heat gained by the water. The entropy change of each would be M C ln (Tfinal/Tinitial)
The water would gain more entropy than the nail loses, for a net gain of the system.
We did a lab in class where we heated a nail, and our goal was to determine the total increase in entropy of the system. I have all of the measurements but I'm not sure how to go about figuring out the change in heat (delta Q).
delta S = delta Q / delta T (S=Q/T)
Here are all of the measurements. If someone could guide me through what to do that would be great.
nail mass = 9.2g
water mass = 185.1g
T initial = 20 degrees Celsius
T final = 22 degrees Celsius
T nail final = 387.8 degrees Celsius
c specific heat nail = 0.11 cal/gC
c specific heat water = 1.00 cal/gC
Thanks.
5 answers
Then maybe I calculated the final temperature of the nail incorrectly. 387.8 degrees Celsius is a value I derived assuming the nail was iron/steel. I used mcT=mcT and solved for the final temperature of the nail using delta T.
Disregarding that possible error, how would I go about calculating the total increase in entropy of the entire system?
Disregarding that possible error, how would I go about calculating the total increase in entropy of the entire system?
Maybe you can't. You need accurate initial and final temperatures of the nail. You don't appear to have them.
Then maybe you could help me attain them.... No initial temperature of the nail was given, and our teacher said that variable wasn't necessary for any step in the experiment, including finding the total change in entropy of the system.
From what you are telling me, I assume that you took an even hotter nail than 387.8 C , immersed it in water, and removed it when the nail's temp was 387.8 C, and the water had risen in T by 2 C. The water temperature rise should have been measured more accurately because you are not going to have a very accurate measurement for the amount of heat transferred. In that case, compute the initial temperature of the nail from the amount of heat transferred, which you can compute from the water temperature rise.
I have already given you the correct formula to use for the entropy change of both water and nail.
delta S = Integral of dQ/T = m C ln(T2/T1)
It involves the natural log (ln) of the absolute temperature ratio. The formula you are using,
delta S = delta Q / delta T
is totally wrong. It should be
delta S = [Q/T]water +
[Q/T]nail
The T's should be average values, in K units, if you are going to use your approximate formula. The Q's, which are the amounts of heat transferred, are equal but of opposite sign. The "delta" is not needed. You can get the Q from the temperature rise of the water.
When you use either formula, the temperatures must be in K. The nail loses entropy, but the water gains more than the nail loses, because of the lower average T in the denominator for water.
I have already given you the correct formula to use for the entropy change of both water and nail.
delta S = Integral of dQ/T = m C ln(T2/T1)
It involves the natural log (ln) of the absolute temperature ratio. The formula you are using,
delta S = delta Q / delta T
is totally wrong. It should be
delta S = [Q/T]water +
[Q/T]nail
The T's should be average values, in K units, if you are going to use your approximate formula. The Q's, which are the amounts of heat transferred, are equal but of opposite sign. The "delta" is not needed. You can get the Q from the temperature rise of the water.
When you use either formula, the temperatures must be in K. The nail loses entropy, but the water gains more than the nail loses, because of the lower average T in the denominator for water.
2 answers