If 50 mL of 10 degrees C water is added to 40 mL of 65 degrees , calculate the final temperature of the mixture assuming no heat is lost to the surroundings, including the container. Is your answer reasonable?
I feel like this should be simple... is it a specific heat problem? How do you do this? Thanks!
use Q = cm(delta T). since no heat is lost, set the two sides equal to each other with one increasing in temp and the other decreasing. convert volume of water to mass based on density, 1 g/ml. (delta T) should be broken down to Tf - Ti. The Tf values are the same, but Ti values are different
Yes, it is a specific heat problem.
Part of the mixture heats up, the other part cools off, and total energy is conserved.
Just do a weighted average because they are the same substance
(50*10+40*65)/90 = ~34.4
I just did the same math and got 29.4.
50*10 = 500
64*40 = 2600
2600+500 = 3100
3100/90 = 34.44444444
Maybe I'm making a really dumb mistake, but I think that's right
CoachedbyMarden wrote:
I just did the same math and got 29.4.
What math did you do?
50 * 10 + 40 * 65 = 500 + 2600 = 3100
50 + 40 = 90
3100/90 ~ 31 + 3.1 + .31 + .031 ... ~ 34.44
Or, take 40 of the 10C and mix with 40 of the 65C to get 80 of 37.5; from this you can tell that having and 8:1 mix of 37.5 and 10 is not going to bring you down to 29.4 (10 up 19.4 to 29.4 and 80 down 8.1 to 29.4 is clearly not right).
Yes I got 34.4 after I solved it a similar way as well.
How about this one?
In a calorimeter, 50 mL of 60 degrees water is added to 50 mL of 14 degrees water. If the max temperature is determined to be 36 deg C, calculate the heat capacity of the calorimeter.
Is this a conservation of energy problem?
I was thinking since heat capacity = C*T, you take (50 g)(60) + (50 g)(14) = 36(100)
something like that but then there is no heat capacity there... not sure how to solve it?