
This problem set deals with thermal
resistance. 


1. 
Thermal resistance is the opposition to
flow of heat energy. 


2. 
A defining equation for thermal
resistance is R_{T} = DT/Q_{H}.



3. 
What is the thermal resistance if you
know the thermal conductivity, k,
the thickness, L, and the surface area, A? (Note: the
capital "L" should be lowercase and cursive)
R_{T} = L/(k·A)



4. 
The thermal conductivity for a given
material is a constant value. It depends on (b.)
type of material. 


5. 
Thermal resistance in English units may
be expressed as F°/(BTU/hr). 


6. 
A building has a thermal resistance of
0.05 C°/(kcal/hr). What is the building's heat flow rate
when the temperature difference between inside and
outside is 30 C°? 

Knowns 
Unknowns 
Equations/Remarks 
R_{T} = 0.05
C°/(kcal/hr)
DT = 30 C° 
Q_{H} = ? 
R_{T} = DT/Q_{H} 
R_{T} = DT/Q_{H} Q_{H} = DT/R_{T} = (30
C°)/(0.05
C°/(kcal/hr))
Q_{H} = 600
kcal/hr

Rearrange the equation to solve
for the heat flowrate, then substitute in the
known values and calculate the answer. 


7. 
In the equation for heat flow rate, Q_{H} = kADT/L, increasing
the thermal conductivity, area, or temperature difference
will increase heat flow rate. (Note: the capital
"L" should be lowercase and cursive) 


8. 
In the equation for heat flow rate, Q_{H} = kADT/L,
increasing the thickness will decrease the
heat flow rate. (Note: the capital "L"
should be lowercase and cursive) 


9. 
The higher the Rvalue rating for a
thermal material is, the stronger its
insulating effect is. 


10. 
Find the thermal resistance, R_{T},
of the plate, given that it is 0.25 inches thick, has an
area of 2 square feet, and is made of copper. Use the
thermal conductivity given in Table II1.


Knowns 
Unknowns 
Equations/Remarks 
L = 0.25 in
A = 2 ft^{2}
k = 2784
(BTU·in)/(hr·ft^{2}·F°) 
R_{T} = ? 
R_{T} = L/(k·A) 
R_{T} = L/(k·A) = (0.25 in)/[(2784
BTU·in/hr·ft^{2}·F°)(2
ft^{2})]
R_{T} = 4.49 ×10^{5}
F°/(BTU/hr)

Substitute in the known values
and calculate the answer. 


11. 
Find the heat flow rate, Q_{H},
through the plate if DT across
the plate is 40 F°. (Refer to problem 10).


Knowns 
Unknowns 
Equations/Remarks 
DT
= 40 F°
R_{T} = 4.49 ×10^{5}
F°/(BTU/hr) 
Q_{H} = ? 
R_{T} = DT/Q_{H} 
R_{T} = DT/Q_{H} Q_{H} = DT/R_{T} = (40
F°)/(4.49 ×10^{5}
F°/(BTU/hr))
Q_{H} = 891,000
BTU/hr

Rearrange the equation to solve
for the heat flowrate, then substitute in the
known values and calculate the answer. 


12. 
The temperature difference between the
inside and outside of an electric oven is 300 F°.
The heating element must supply 1200 BTU/hr of heat
energy to the oven to overcome the heat flow out of the
oven. Find the thermal resistance of the oven walls.


Knowns 
Unknowns 
Equations/Remarks 
DT
= 300 F°
Q_{H} = 1200 BTU/hr 
R_{T} = ? 
R_{T} = DT/Q_{H} 
R_{T} = DT/Q_{H} = (300 F°)/(1200
BTU/hr) R_{T} = 0.25
F°/(BTU/hr)

Substitute in the known values
and calculate the answer. 


13. 
The thermal resistance of a plate of
glass is 0.006 F°/(BTU/hr). 9600 BTU of heat energy flow
through the plate glass every hour. Find the temperature
difference across the glass that is causing the flow of
heat energy.


Knowns 
Unknowns 
Equations/Remarks 
R_{T} =
0.006 F°/(BTU/hr)
Q_{H} = 9600 BTU/hr 
DT
= ? 
R_{T} = DT/Q_{H} 
R_{T} = DT/Q_{H} DT = R_{T} · Q_{H} = (0.006
F°/BTU/hr)(9600
BTU/hr)
DT = 57.6
F°

Rearrange the equation to solve
for the temperature difference, substitute in the
known values and calculate the answer. 


14. 
The thermal resistance of a 16 ft^{2}
piece of 0.5 inch thick aluminum is 19.1 ×10^{6}
F°/(BTU/hr). Find the thermal conductivity for the
aluminum plate.


Knowns 
Unknowns 
Equations/Remarks 
R_{T} =
19.1 ×10^{6} F°/(BTU/hr)
L = 0.5 in
A = 16 ft^{2} 
k
= ? 
R_{T} = L/(k·A) 
R_{T} = L/(k·A) k = L/(R_{T}·A) = (0.5
in)/[(19.1 ×10^{6} F°/BTU/hr)(16
ft^{2})]
k = 1636
(BTU·in)/(hr·ft^{2}·F°)

Rearrange the equation to solve
for the thermal conductivity, substitute in the
known values and calculate the answer. 


15. 
A firefighter's suit has a thermal
resistance of 50 C°/(kcal/hr). When exposed to a fire,
the temperature difference from outside to inside the
suit may be as high as 1000 C°. Find the heat flow
rate through the suit at 1000 C°.


Knowns 
Unknowns 
Equations/Remarks 
DT
= 1000 C°
R_{T} = 50 C°/(kcal/hr) 
Q_{H} = 
R_{T} = DT/Q_{H} 
R_{T} = DT/Q_{H} Q_{H} = DT/R_{T} = (1000
C°)/(50
C°/kcal/hr)
Q_{H} = 20
kcal/hr

Rearrange the equation to solve
for the heat flowrate, then substitute in the
known values and calculate the answer. 


16. 
Given the conditions of problem 14 and a
constant temperature difference of 50 F° across the
material,
 find the heat flow rate using the equation Q_{H} = kADT/L,
(Note: the capital "L" should be
lowercase and cursive)
 Is there an easier way to do this? If so solve
for Q_{H} the easiest way.


Knowns 
Unknowns 
Equations/Remarks 
R_{T} =
19.1 ×10^{6} F°/(BTU/hr)
L = 0.5 in
A = 16 ft^{2}
DT = 50 F°
k = 1636
(BTU·in)/(hr·ft^{2}·F°) 
a. Q_{H} = ?
b. Q_{H} = ? by the easiest way 
Q_{H} = kADT/L
R_{T} = DT/Q_{H}

a. Q_{H} = kADT/L
Q_{H} = (1636
BTU·in/hr·ft^{2}·F°)(16
ft^{2})(50
F°)/(0.5
in)
Q_{H} = 2,620,000
BTU/hr b. R_{T} = DT/Q_{H}
Q_{H} = DT/R_{T} = (50
F°)/(19.1 ×10^{6}
F°/BTU/hr)
Q_{H} = 2,620,000
BTU/hr

Substitute in the known values
and calculate the answer. The easier way is to
use the thermal resistance equation.



17. 
 Using the table of thermal conductivity (Table
II1), find the thermal resistance of a
36" x 40" windowpane which is
0.125 in thick.
 If the inside of the window is at 70 °F and
the outside window surface is at 65 °F,
find the heat flow rate through the window.
 How many BTU's will escape through the window in
8 hours?


Knowns 
Unknowns 
Equations/Remarks 
k = 5.8
(BTU·in)/(hr·ft^{2}·F°)
W = 36 in
H = 40 in
L = 0.125 in
T_{1} = 70 °F
T_{2} = 65 °F
t = 8 hrs 
a. R_{T} = ?
b. Q_{H} = ?
c. H = ? 
R_{T} = L/(k·A)
A = W·H
DT = T_{1} – T_{2}
R_{T} = DT/Q_{H}
Q_{H} = H/t 
a. A
= W·H = (36 in)(40
in)(1
ft^{2}/144 in^{2}) = 10
ft^{2}
R_{T} = L/(k·A) = (0.125 in)/[(5.8
BTU·in/hr·ft^{2}·F°)(10
ft^{2})]
R_{T} = 2.16 ×10^{3}
F°/(BTU/hr) b. DT = T_{1} – T_{2} = 70 °F – 65 °F = 5 F°
R_{T} = DT/Q_{H}
Q_{H} = DT/R_{T} = (5
F°)/(2.155 ×10^{3}
F°/BTU/hr)
Q_{H} = 2320
BTU/hr
c. Q_{H} = H/t
H = Q_{H}·t = (2320
BTU/hr)(8
hr)
H = 18,600 BTU

Find the area of the window and
use it along with the thickness and thermal
conductivity to find the thermal resistance.
Notice that the results do not match those in the
book. The possible cause could be a different k. Find the temperature
difference across the window, then use it with
the thermal resistance to find the heat flowrate.
Solve the next equation for H, substitute in
the numbers, and calculate the answer.
Note that all three answers differ from those
in the book.



18. 
A container is filled with 800
milliliters of water at 95 °C. After sitting for 40
minutes in a room whose temperature is 25 °C, the
water has cooled to 80 °C.
 Find the temperature change of the water, D_{T} = T_{f}  T_{i}.
What is the meaning of a negative temperature
change?
 How many calories did the water lose?
 Find the heat flow rate in cal/sec.
 What was the average temperature of the water as
it cooled?
 What was the temperature difference between the
room and the average water temperature? (DT = T_{av}  T_{room})
 Find the thermal resistance of the container.
 Explain the difference in meanings for the DT in part a and part e
above.


Knowns 
Unknowns 
Equations/Remarks 
V = 800 mL
t = 40 min
T_{i} = 95 °C
T_{f} = 80 °C
T_{room} = 25 °C
c = 1.0 cal/g·C°
r = 1.0 g/cm^{3}
1 cm^{3} = 1 mL
1 min = 60 sec 
a. DT = ?
b. H = ?
c. Q_{H} = ?
d. T_{av} = ?
e. DT = ?
f. R_{T} = ?
g. Explain the different DT's 
D_{T} = T_{f} – T_{i}
r = m/V
H = mcDT
Q_{H} = H/t
T_{av} = (T_{f} + T_{i})/2
DT = T_{av}  T_{room}
R_{T} = DT/Q_{H}

a. D_{T} = T_{f} – T_{i} = 80 °C – 95 °C = –15 C°.
The negative sign means that the water cooled
down. b. r = m/V
m = rV
H = mcDT = rVcDT
H = (1.0 g/cm^{3})(800
cm^{3})(1.0
cal/g·C°)(–15 C°)
H = 12,000 cal
c. Q_{H} = H/t = (12000
cal)/[(40 min)(60
sec/1 min)]
Q_{H} = 5
cal/s
d. T_{av} = (T_{f} + T_{i})/2 = (80 °C + 95 °C)/2
T_{av} = 87.5 °C
e. DT = T_{av}  T_{room} = 87.5 °C – 25 °C
DT = 62.5 C°
f. R_{T} = DT/Q_{H} = (62.5 C°)/(5
cal/s)
R_{T} = 12.5
C°/(cal/s)
g. The DT
in part (a) is the change in
temperature of the water (one
substance). The DT
in part (e) is the difference between
the water and the air in the room (two
substances). It is this DT
that drives the flow of heat.

Find the change in temperature
of the water. Use the density equation to find
the mass of the water from the given volume and
subsitute that in the heat equation. Use the
absolute value of the temperature difference
since we know that the water lost the heat.
For the next few parts, substitute in the
known values and calculate the answers. Convert
the minutes to seconds.



19. 
Using the same process as above, find
the thermal resistance of a container which holds 5
gallons of water at 200 °F if the water cools to
170 °F in 15 minutes. The room temperature is
75 °F.


Knowns 
Unknowns 
Equations/Remarks 
V = 5 gal
t = 15 min
T_{i} = 200 °F
T_{f} = 170 °F
T_{room} = 75 °F
c = 1.0 BTU/lb·F°
r_{w} = 62.4
lb/ft^{3}
1 ft^{3} = 7.48 gal
1 hr = 60 min

R_{T} = ? 
D_{T} = T_{f} – T_{i}
r_{w} = w/V
H = wcDT
Q_{H} = H/t
T_{av} = (T_{f} + T_{i})/2
DT = T_{av}  T_{room}
R_{T} = DT/Q_{H} 
D_{T} = T_{f} – T_{i} = 170 °F – 200 °F = –30 F° r_{w} = w/V
w = r_{w}·V
H = wcDT = r_{w}VcDT
H = (62.4 lb/ft^{3})(5
gal)(1
ft^{3}/7.48
gal)(1.0
BTU/lb·F°)(–30 F°)
H = 1251 BTU
Q_{H} = H/t = (1251
BTU)/[(15 min)(1
hr/60 min)]
Q_{H} = 5004 BTU/hr
T_{av} = (T_{f} + T_{i})/2 = (170 °F + 200 °F)/2
T_{av} = 185 °F
DT = T_{av}  T_{room} = 185 °F – 75 °F
DT = 110 F°
R_{T} = DT/Q_{H} = (110 F°)/(5004
BTU/hr)
R_{T} = 0.0220
F°/(BTU/hr)

Find the change in water
temperature (negative means the water cooled
down). Find the amount of heat transferred out
of the water. Use the absolute value of the
temperature change since we know that the heat is
going out of the water.
Go through the next few steps, substituting in
the known values and calculating the results.
Convert minutes to hours since BTU/hr is a common
unit.
Finally we get to the end result.

