## 20141213

### Physics quiz question: comparing changes in internal energies

Physics 205A Quiz 7, fall semester 2014
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e Conceptual Question 14.1, Practice Problem 14.5, Comprehensive Problem 14.75

"French Press Coffee II"
flic.kr/p/57FhdB

0.040 kg of coffee grounds at 25.0° C is placed into a French press container with 1.4 kg of 90.0° C water. The coffee grounds and water are allowed to reach thermal equilibrium. Ignore the effects of evaporation and phase changes, and heat exchanged with the environment or container. Specific heat of coffee grounds[*] is 1,700 J/(kg·K). Specific heat of water is 4,190 J/(kg·K). After reaching thermal equilibrium, the __________ had the greatest change in internal energy.
(A) coffee grounds.
(B) water.
(C) (There is a tie.)
(D) (Not enough information is given.)

[*] Hans Dieter Baehr, Karl Stephan, Heat and Mass Transfer (2nd ed.), Springer-Verlag (2006), p. 368.

Correct answer (highlight to unhide): (C)

The transfer/balance energy conservation equation for this system is given by:

Qext = ∆Ecoffee + ∆Ewater,

where for an isolated system, Qext = 0, such that:

0 = mcoffee·ccoffee·(Tcoffee, f – 25.0° C) + mwater·cwater·(Twater, f – 90.0° C),

which can be subsequently solved for the final equilibrium temperature (Tcoffee, f = Twater, f = 89.255139872° C, or 89° C to two significant figures. While this means that the coffee experienced a larger temperature change than the water, the coffee experienced the same amount of thermal internal energy change (an increase) as the change in thermal internal energy of the water (a decrease). This can be seen by looking back on the transfer/balance energy conservation equation:

0 = ∆Ecoffee + ∆Ewater,

–∆Ecoffee = ∆Ewater.

Sections 70854, 70855, 73320
Exam code: quiz07cO4t
(A) : 12 students
(B) : 7 students
(C) : 45 students
(D) : 0 students

Success level: 70%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.53