Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Consider you have three identical, well insulated containers. Container A has 1

ID: 1473769 • Letter: C

Question

Consider you have three identical, well insulated containers. Container A has 1 kg of water at 80oC, the container B has 1 kg of water at 20oC, and the third container C is empty. You have a fourth container, D; with excellent heat conducting walls and which can easily fit inside any of the other three. You are not allowed to mix the water in the containers, but you can move water from one to another container. How can you heat the water in container B using the water in container A (by cooling it, of course) with the aid of container D (and C) to a final temperature higher than the final temperature in container A. (At the end of the procedure, the temperature of the water in A may not be 80oC). State the temperatures in each container. Assume that there is no energy or material loss during the procedure you describe, and you are working in a vacuum, i.e. no heat loss or gain to the environment.

Explanation / Answer

Given

Mass of water in container A mA = 1 kg

Mass of water in container B mB = 1 kg

The specific heat capacity of water c = 4190 J/kgoC

Temperature in container A TA = 80oC

Temperature in container B TB = 20oC

Solution

Heat energy in container A QA = mACTA

QA = 1 x 4190 x 80

QA = 335200 J

Heat energy in container B QB = mBCTB

QB = 1 x 4190 x20

QB = 83800 J

If we transfer half of water in container B to D and place the container D inside A

Heat lost by container A = Heat gained by container D

mAC(TA –T) = mBC(T-TB)/2

since mA = mB = 1 kg

2TA – 2T = T - TB

3T = 2TA + TB

T = 60oC

Now container A and D are at 60oC

now we transfer contents of D into C and pour the other half of container B into D and we place D into A

now heat lost by A = heat gained by D

mAC(T –T’) = mBC(T’-TB)/2

2T – 2T’ = T’ - TB

T’ = 46.67OC

Now we transfer contents of D and C back to B

the new temperature is T1

mBC(T –T1)/2 = mBC(T1 – T’)

2T1 = T + T’

T1 = 53.34oC

Finally, now we can see the temperature of container B is greater than the temperature of container A

53.34oC > 46.67OC

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote