ENERGY USE AT HOME
Home Heating
Problem Set
Problem 1A: Heating a Wooden Cabin
A windowless wooden cabin, with a well insulated floor and roof, has walls 15 cm thick, 10 m long, 10 m
wide and 3 m high. If the temperature outside is 5oC, will a 1000 W space heater be enough to keep the
cabin at a comfortable temperature? (kwood = 0.10 W/mK or 0.10 W/moC)
Problem 1A Solution: Heating a Wooden Cabin
The house is in dynamic equilibrium so conservation of energy tells us that the flow of heat through the
walls has to be equal to the power generated by the space heater.
Pin = Pout
The heat lost by conduction is given by:
Q
kA(Thot  Tcold )
x
Where k is the thermal conductivity of the wood, A is the area over which heat is being lost, Tcold is the
temperature outside, Thot is the temperature inside, and x is the thickness of the wood.
We can rearrange this equation to give us Thot.
Thot 
Qx
 Tcold
kA
Since we know that:
Q = 1000 W
x = 0.15 m
k = 0.10 W/moC
A = 10 m * 3 m * 4 sides = 120 m2
Tcold = 5oC
We can find the temperature inside the cabin:
Qx
 Tcold
kA
(1000 W)(0.15 m)

 5o C
o
2
(0.10 W/m C)(120 m )
Thot 
 12.5 o C  5 o C
 17.5 o C
©University of British Columbia Department of Physics and Astronomy
So, the temperature inside the cabin is 17.5oC. Note, a real log cabin would have leaky windows and
doors and so the heat loss would be considerably more and temperature inside would be much lower.
Problem 1B: Heating a Wooden Cabin
Two campers, who have consumed about 2200 Cal each today, spend their evening relaxing around the
space heater. If all the calories are burned evenly in a 24 hour period, how much will their body heat
contribute to the temperature inside the cabin?
Problem 1B Solution: Heating a Wooden Cabin
We know that 1 Cal (1 food calorie = 1kcal) is equal to 4200 J (see Useful Numbers article). So in one
day they have consumed:
2200 Cal  2200 kcal (
4200 J
)  9.24x10 6 J
1 kcal
There are 86 400 seconds in a day. Knowing this we can calculate how much power this is equivalent to:
9.24x10 6 J
 107 W
86400 s
The human body produces power (for a man at rest) at a rate of 100 W (this can be altered by activity
level). If more food is consumed than this, the calories get stored as fat and if less food is consumed,
stored calories are burned, so in both cases the output remains similar. In this case, each camper
produces power at a rate of about 100 W, and consumed about the same amount of calories as they
burned. Adding this to the power produced by the space heater, we can calculate how this affects the
temperature in the cabin.
Qx
 Tcold
kA
(1200 W)(0.15 m)

 5o C
(0.10 W/m o C)(120 m 2 )
Thot 
 15 o C  5 o C
 20 o C
The temperature inside the cabin is now 20oC which a comfortable ambient temperature, but again, in a
real cabin the temperature would be lower due to the leaky doors and windows.
©University of British Columbia Department of Physics and Astronomy
Problem 2: Insulation
A builder is going to build a house (10 m by 10 m and 3 m tall) with brick that is 4 inches thick and has an
R-value of R-4. In an attempt to save money and energy, he only plans to keep his house heated from
October to April where he will keep the temperature approximately 20oC warmer than it is outside. He
is trying to decide whether or not he should insulate the exterior walls of his brick house. After a trip to
his local hardware store, he finds that for approximately $55, he could insulate 75 ft2 with fibreglass
insulation (R-20). Compare the predicted heat loss for his house with and without insulation and
determine if it is worth it to insulate the exterior walls of his house
Problem 2: Insulation Solution
The cost of insulation:
The area of his house will be 10 m x 3 m x 4 sides = 120 m2
Cost of buying insulation per m2:
$55
1 ft
(
Total insulation costs:
(
Heat Lost:
If there is only brick:
75 ft
2
)(
0.3048 m
)2 
$7.89
m2
$7.89
) 120 m 2  $950
2
m
Q AT
(120 m 2 )( 20 K)


 3400 W
T
R
(4 R)(0.176 m 2 K / WR )
If the walls are insulated:
Q AT
(120 m 2 )( 20 K)


 570 W
T
R
(4 R  20 R)(0.176 m 2 K / WR )
The cost of electricity each year:
In Vancouver, electricity costs $11/GJ * and our builder is only going to keep his house heated
for 7 months each year.
If there is only brick:
J
30 days 24 h 60 min 60 s
 (3400 )(7 months)(
)(
)(
)(
)
s
1 month 1 day
1h
1 min
$11
 6.17x10 10 J(
)
1x10 9 J
 $700
If the walls are insulated:
J
30 days 24 h 60 min 60 s
 (570 )(7 months)(
)(
)(
)(
)
s
1 month 1 day
1h
1 min
$11
 1.03x10 10 J(
)
1x10 9 J
 $100
©University of British Columbia Department of Physics and Astronomy
So, the builder will save approximately $600 a year if he insulates his house and so his savings will pay
off the cost of the insulation in less than 2 years.
* Terasen Gas. Rates (online). http://www.terasengas.com/Homes/Rates/default.htm?region=all [7
July 2010].
Problem 3: Radiation
What happens to the heat after it has been conducted through the wall? Heat that is conducted to
exterior walls is carried away by convection and radiation. On a winter’s evening with a low overcast
sky, the cabin’s environment is equivalent to being surrounded by a black body of temperature 0oC.
For the same house as problem 3, if the temperature of the surface of the house is 5oC, and it has an
emissivity of 1.0, how much heat is lost due to radiation?
Problem 3 Solution: Radiation
We know:
e = 1.0
σ = 5.67 x 10-8 W/m2K4
A = 120 m2
T2 = 11oC = 273 K
T1 = 12oC = 278 K
Heat lost to radiation:
Q
 eA(T24  T14 )
T
 (1.0)(5.67 x 10 -8 W/m 2 K 4 )(120 m 2 )((278 K) 4  (273K) 4 )
 2,850 W
We can see from this that radiation is a very effective way to get rid of heat once it has been conducted
through the wall.
Problem 4: Laptop in the Sun
On a warm summer day, you leave your laptop computer out in plain sunlight for a few hours. The black
plastic lid (A = 0.092 m2) is closed and is exposed to the sunlight.
A) What is the total power on the lid due to the sun’s radiation?
B) How much power is radiated by the lid at the equilibrium temperature?
C) What is the surface equilibrium temperature of the lid?
©University of British Columbia Department of Physics and Astronomy
Problem 4 Solution: Laptop in the Sun
A) What is the total power on the lid due to the sun’s radiation?
The solar constant describes the power of the sun per m2, S = 1367 W/m2. The lid absorbs:
P  (1367 W/m 2 )(0.092 m 2 )  126 W
B) How much power is radiated by the lid at the equilibrium temperature?
At equilibrium temperature Pin = Pout. The incoming power is the one calculated in (A) and the
outgoing power is due to the temperature of the lid due to the Stefan Boltzmann law.
Pout  eAT 4  126 W
C) What is the surface equilibrium temperature of the lid?
In this case there is no To since the environmental temperature is mainly radiated by nearby
objects and if the laptop is on an open area, nothing can radiate back to the top of the lid.
Pout  eσeσ
T4 
4
 126 W
126 W
eσσ
126 W
(1)(5.67x1 0 W/m 2 K)(0.092 m 2 )
 394 W

8
Problem 5: The Heating of a House
For the following question, use the dimensions and characteristics of your own house. It is reasonable
to assume that walls are thick enough to block direct heat loss to radiations (without prior conduction
through the walls). Windows are relatively thin in comparison and conduct more heat than walls (i.e. a
double-pane glazed window has a typical value of R-2).
A) How much heat is lost due to conduction?
B) How much heat is radiated directly (without conduction)?
C) How much heat is lost through convection (you can assume the air is completely exchanged 3
times / day if you don’t know this value for your own house).
D) What is the total heat loss? What are the shortcomings of your model?
E) Electrical energy is typically generated in coal-fired power plants which emit approximately 1.55
lbs (0.703 kg) of CO2 per kWh produced*. How much money would heating your house require
and how much CO2 would it produce each day? Each year?
F) How much money and CO2 could you save by lowering the temperature by 1oC?
*Energy Star. Life Cycle Cost Estimate for ENERGY STAR Qualified Dehumidifiers Assumptions (online).
http://www.energystar.gov/ia/business/bulk_purchasing/bpsavings_calc/CalculatorConsumerDehumidi
fier.xls [7 July 2010].
©University of British Columbia Department of Physics and Astronomy
Problem 5 Solution: The Heating of a House
For the following question, use the dimensions and characteristics of your own house. It is reasonable
to assume that walls are thick enough to block direct heat loss to radiations (without prior conduction
through the walls). Windows are relatively thin in comparison and conduct more heat than walls (i.e. a
double-pane glazed window has a typical value of R-2).
For our answers we will use a house (5 m tall) with 10 m2 of windows (R-2), 90 m2 of walls (R-20), and an
average temperature difference of 10oC (20oC inside, 10oC outside).
A)
How much heat is lost due to conduction?
Walls:
A = 90 m2
𝛥T = 10 K
R-20
Windows:
A = 10 m2
𝛥T = 10 K
R-2
Q AT
(90 m 2 )(10 K)


 256 W
T
R
(20 R)(0.176 m 2 K / WR )
Q AT
(10 m 2 )(10 K)


 284 W
T
R
(2 R)(0.176 m 2 K / WR )
Overall, 540 W are lost to conduction.
B)
How much heat is radiated directly (without conduction)?
Recall that the walls are thick enough to block direct heat loss due to radiation.
Windows:
Q
 eA(T24  T14 )
e = 1.0
T
A = 10 m2
 (1.0)(5.67 x 10 -8 W/m 2 K 4 )(10 m 2 )((293 K) 4  (283 K) 4 )
o
Tin = 293 K (20 C)
 542 W
Tout = 283 (10oC)
But, only about half of this heat is lost because the other half is radiated back into the room
(remember the basis of the greenhouse effect). So, the heat loss to radiation is 270 W.
C)
How much heat is lost through convection (you can assume the air is completely
exchanged 3 times / day if you don’t know this value for your own house).
Volume = (90 m2 + 10 m2)(5 m) = 500 m3
Volume of air exchanged each day = (500 m3)(3) = 1500 m3
𝛥T = 10 K
Specific Heat Capacity of Air, c = 1005 KJ/kg⦁K
Density of air, ρ = 1.2 kg/m3
©University of British Columbia Department of Physics and Astronomy
Heat, Q  cTV
 (1005
KJ
kg
)(10 K)(1.2 3 )(1500 m 3 )
kgK
m
 18 MJ
Power 
Q

t
18 MJ
 208 W
min
s
(24 h)(60
)(60
)
h
min
Through convection, 210 W are lost.
D)
What is the total heat loss? What are the shortcomings of your model?
Thermal energy lost:
Conduction: 540 W
Radiation:
270 W
Convection: + 210 W
Total:
1020 W
In this model, we did not account for ventilation, cracks or the opening of doors or windows and
we also ignored heat loss through the floor and ceiling.
E)
Electrical energy is typically generated in coal-fired power plants which emit
approximately 1.55 lbs (0.703 kg) of CO2 per kWh produces*. How much money would
heating your house require and how much CO2 would it produce each day? Each year?
Terasen Gas charges ~$11/GJ *
Each day:
Energy  (Power) (time)
min
s
)(60
)
h
min
 8.8 x 10 7  0.088 GJ
CO 2 : (1.02 kW)(24 h)(
0.703 kg
)  17.2 kg
kWh
 (1020 W)( 24 h)(60
Cost  (Rate) (Energy)
(
$11
)(0.088 GJ )  $0.97
1 GJ
Each year:
Cost ≈ $355, CO2: 6200 kg
* Terasen Gas. Rates (online). http://www.terasengas.com/Homes/Rates/default.htm?region=all [7
July 2010].
©University of British Columbia Department of Physics and Astronomy
F)
How much money and CO2 could you save by lowering the temperature inside by 1oC?
(we will lower it to 19oC).
Conduction:
Walls:
Q AT
(90 m 2 )(9 K)


 230 W
T
R
(20 R)(0.176 m 2 K / WR )
Windows:
Q AT
(10 m 2 )(9 K)


 256 W
T
R
(2 R)(0.176 m 2 K / WR )
Radiation:
Q
 eA(T24  T14 )
T
 (1.0)(5.67 x 10 -8 W/m 2 K 4 )(10 m 2 )((292 K) 4  (283 K) 4 )
 485 W
 243 W (because half of it is radiated back into the house)
Convection:
Q
cTV

ΔT
t
kJ
kg
)(9 K)(1.2 3 )(1500 m 3 )
kgK
m
min
s
(24 h)(60
)(60
)
h
min
(1005

 188 W
Thermal energy lost:
Conduction: 486 W
Radiation:
243 W
Convection: + 188 W
Total:
917 W
Cost  (Rate) (Energy)
(
$11
min
s
1 GJ
)(920 W)( 24 h)(60
)(60
)(
)  $0.87
1 GJ
h
min 1 x 10 9 J
CO 2 : (0.92 kW)(24 h)(
0.703 kg
)  15.5 kg
kWh
Savings:
Cost: $0.10/day – $36 / year
CO2: 1.7 kg/day – 640 kg/year
Andrzej Kotlicki, Georg Rieger and Brittany Tymos 2010-07-15
©University of British Columbia Department of Physics and Astronomy
©University of British Columbia Department of Physics and Astronomy