Phys211 Mp 03

Front 1

Consider the following statements:
A. The entropy of an isolated system never decreases.
B. Heat never flows spontaneously from cold to hot.
C. The total thermal energy of an isolated system is constant.
Which of these express the second law of thermodynamics?

Back 1

A, B

Front 2

A heat engine is designed to do work. This is possible only if certain relationships between the heats and temperatures at the input and output hold true.

If Q_h is the magnitude of heat absorbed, and Q_c is the magnitude of heat liberated, what must be true of the relationships between Q_h and Q_c and T_h and T_c for a heat engine to do work?

Back 2

Q_h > Q_c and T_h > T_c

Front 3

Express W done by an ideal heat engine in terms of Q_h and Q_c.

Back 3

W = Q_h - Q_c

Front 4

Express thermal efficiency (e) of a heat engine in terms of Q_h and Q_c

Back 4

e = (Q_h - Q_c) / Q_h

Front 5

Given heat engines:

A. P_in=1000W, T_h = 600K, P_out=600W, T_c=300K, Output Power = 400 W

B. P_in=1000W, T_h = 600K, P_out=0W, T_c=300K, Output Power = 1000 W

C. P_in=1000W, T_h = 600K, P_out=600W, T_c=200K, Output Power = 600 W

D. P_in=1000W, T_h = 500K, P_out=500W, T_c=300K, Output Power = 500 W

E. P_in=1000W, T_h = 500K, P_out=550W, T_c=250K, Output Power = 450 W

F. P_in=1000W, T_h = 600K, P_out=250W, T_c=200K, Output Power = 650 W

Which of these violates the first law of thermodynamics?

Of those remaining, which violate the second law of thermodynamics?

Which of the remaining designs has the highest thermal efficiency?

Back 5

C, F
∆E_th = W + Q
W is Output Power, Q is power in, ∆E_th is power out

B,D
Find e_max using Carnot's 1 - (T_c / T_h), then find e using W_out / Q_h. if e > e_max, the second law has been violated.

E
.45 compared to A's .4

Front 6

Carnot's Equation

Back 6

calculates max efficiency

e_max = 1 - (T_c / T_h)

where T is the Kelvin temperature of a cold and a hot reservoir

Front 7

While keeping your food cold, your refrigerator transfers energy from the inside to the surroundings. Thus, thermal energy goes from a colder object to a warmer one.

Does this violate the 2nd law of thermodynamics? Why or why not?

Back 7

The second law of thermodynamics applies in this situation, but it is not violated because the energy did not spontaneously go from cold to hot.

Front 8

Determine the efficiency of a heat engine where Q_h = 100 J, Q_c = 60 J, and W = 40 J

Back 8

e = W / Q_h = 0.4

Front 9

Relate Q_h, Q_c, and W_in for a heat pump.

Back 9

Q_h = Q_c + W_in

Front 10

COP

Back 10

coefficient of performance

calculation depends on whether the heat pump is used for cooling (refrigerator) or for heating (geothermal heating system)

Front 11

COP for heat pump that cools

Back 11

e.g. a refrigerator

COP = Q_c / W_in = what you get / what you paid

Front 12

COP_max for heat pump that cools

Back 12

e.g. a refrigerator

COP_max = T_c / (T_h - T_c)

Front 13

COP for heat pump that warms

Back 13

e.g. a geothermal heating system

COP = Q_h / W_in = what you get / what you paid

Front 14

COP_max for pump that heats

Back 14

e.g. a geothermal heating system

COP_max = T_h / (T_h - T_c)

Front 15

COP & COP_max for heating vs cooling

Back 15

COP = what you get / what you paid
for cooling, what you get = Q_c
for heating, what you get = Q_h
either way, what you paid is W_in

COP_max = temp you want / (temp of hot reservoir - temp of cool reservoir)
for cooling, temp you want is T_c
for heating, temp you want is T_h

Front 16

Calculate the coefficient of performance for a refrigerator that takes in 40 J of work and exhausts 100 J of heat.

Back 16

Q_c = Q_h - W_in = 60 J

COP = Q_c / W_in = 1.5