Chapter 13
A highway is made of concrete slabs that are 15 m long at
20C. The temperature range for the road is -20 to +40C.
a) What is the minimum size of the expansion gap at 20C
DL = L a DT
= (15) (12x10-6) (20)
= 3.6 mm
b) What is the size of the expansion gap when it is -20C?
DL = L a DT
= (15) (12x10-6) (60)
= 10.8 mm
Chapter 13
An ideal gas that occupies 1.2m3 at a pressure of 105 Pa
and a temperature of 27C is compressed to 0.6 m3 and
heated to 227C. What is the new pressure?
PV = n R T
P1 V1 / T1 = n R
P2 V2 / T2 = n R
P1 V1 / T1 = P2 V2 / T2
P2 = P1 ( V1/V2) (T2/T1)
= 105 ( 1.2/0.6) ( 273+227)/(273+27)
= 3.3 x 105
Chapter 13
What is the temperature of an ideal gas whose molecules
in random motion have an average translational internal
energy of 3.20x10-20 J
U = (3/2) k T
T = (2/3) U/k
= (2/3) (3.2x10-20)/(1.38x10-23)
= 1546 K
Chapter 14
How much heat is required to change 1 Kg of ice at -20C
into steam at 110C. (note Cice = 0.5 kcal/(Kg K) CH20 = 1
kcal/(Kg K) Lf = 79 kcal/Kg, Lv = 539 kcal/Kg)
Q = Cice DTiceM + Lf M + Cwater DTwater M + LV M
+ CsteamDTsteamM
= (0.5) (0- -20) + 79 + 1 (100-0) + 539 + 0.5(110 – 100)
= 733
Chapter 14
For a temperature difference of 20C, a slab of material
conducts 10 W/m2 ; another of the same shape conducts
20 W/m2. What is the rate of heat flow through both slabs
placed next to each other, with a temperature difference
of 20C.
I = kADT / d
= DT / R
R = DT / I
R1 = DT / I1
= 20/10 = 2 C m2 / W
R2 = DT / I2
= 20/20 = 1 C m2 / W
R12 = R1 + R2 = 3 C m2 / W
I12 = DT / R12
= 20 / 3
= 6.67 W/m2
Chapter 15
A monatomic ideal gas at 27C undergoes a constant
volume process, and then a constant pressure process as
shown. During this it changes from 1 liter at 2 ATM to 2L
at 1 ATM
a) What is the work done by the gas.
P
1
W = PDV
2
W12 = 0
3
W23 = (1x105 )(10-3)
= 100 J
V
b) What is the final temperature of the gas?
PV = n R T
P1 V1 / T1 = n R
P2 V2 / T2 = n R
P1 V1 / T1 = P2 V2 / T2
T2 = T1 ( V2/V1) (T1/T2)
= T1 ( 2/1) (1/2)
= T1
Chapter 15
An engine releases 0.45kJ of heat for every 0.1 kJ of
work it does.
a) What is the efficiency of the engine?
e = W / QH
= W / (W + Qc)
= 0.1 / (0.1 + 0.45)
= 0.18
b) If the waste heat is expelled at room temperature
(300K), what is the minimum temperature TH that the heat
is input?
e < 1 - TC / TH
Carnot is maximum efficiency
Tc/TH < 1-e
Tc/ (1-e) < TH
TH > Tc / (1-e)
> 300 / (1 - .18)
> 366