Physics 202 Homework, Day 03: Chapter 11 : 38, 41, 44, 46, 47

Physics 202 Homework, Day 03: Chapter 11 : 38, 41, 44, 46, 47
ü 38. A glass windowpane in a home is 0.62 cm thick and has dimensions of 1.0m x 2.0 m. On a certain day,
the indoor temperature is 25°C and the outside temperature is 0°C.
(a) What is the rate of heat flow through the glass?
(b) How much energy is lost through the window in one day, assuming the inside and outside
temperatures remain constant ?
In[13]:=
Dx = .0062; k = 0.84; DT = 25 - 0; WindowArea = 2.0; HeatLossPerSecond = k * WindowArea * DT
Dx ;
Print@"The heat loss per second is ", HeatLossPerSecond, " Joules."D
The heat loss per second is 6774.19 Joules.
In[15]:=
Print@" The heat loss per day is ", HeatLossPerSecond * 24 * 60 * 60, " Joules."D
The heat loss per day is 5.8529 µ 10 8
Joules.
ü 41. A steam pipe is covered with 1.50 cm thick insulating material of thermal conductivity 0.200 cal/(cm °C
s). How much energy is lost every second when the steam is at 200°C, and the surrounding air is at
20.0°C? The pipe has a circumference of 800 cm and a length of 50.0m. Neglect losses though the ends of
the pipe.
The units for this problem are calories, centimeters and °C. Since k is given in these units, we will use thickness in cm, and
area in cm 2 , giving an answer in calories. Multiplying this number by 4.186 will give the answer in Joules.
In[43]:= Dx = 1.50; k = 0.20; DT = 200 - 20; PipeArea = 800. * H50 * 100L;
HeatLossPerSecond = k * PipeArea * DT
Dx ;
Print@" The heat loss per second is ", HeatLossPerSecond,
" calories, or ", HeatLossPerSecond * 4.186, " Joules."D
The heat loss per second is 9.6 µ 10 7 calories, or 4.01856 µ 10 8 Joules.
Note: This seems like a *huge* loss of energy, so I checked it and found that the insulating material chosen is *terrible* - it
has the same thermal conductivity as iron! A good insulator, equivalent to the insulating power of styrofoam, would give an
answer about 1200 times less, or about 3.5 x 10 5 Watts - 345 kW, instead of 402 MW. 400MW is comparable to the output
power of an entire electrical power plant! Sounds like time for another note to the author...
ü 44. A thermopane window consists of two glass panes, each 0.50 cm thick, with a 1.0-cm-thick sealed
layer of air in between. If the inside surface temperature is 23.0 °C, and the outside surface temperature is
0.0°C, determine the rate of energy transfer though 1.0 m 2 of the window. Compare your answer with the
rate of energy transfer though 1.0 m 2 of a single 1.0-cm-thick pane of glass.
In[88]:= DxGlass = .0050; kGlass = 0.84; DxAir = .010; kAir = 0.0234; DT = 23 - 0; WindowArea = 1.0;
HeatLossPerSecond =
J DxGlass
kGlass
WindowArea * DT
+ DxAir
kAir
;
+ DxGlass N kGlass
Print@"Heat loss per second through Thermopane glass = ", HeatLossPerSecond, " Watts."D
HeatLossPerSecond1 =
WindowArea * DT
J DxGlass
kGlass
;
+ DxGlass N kGlass
Print@"Heat loss per second through double-thick glass = ", HeatLossPerSecond1, " Watts."D

2 Day_03_sol.nb
Heat loss per second through Thermopane glass = 52.3614 Watts.
Heat loss per second through double-thick glass = 1932. Watts.
That is a *huge* difference - 37 times more heat loss without the layer of air in between.
ü 46. A Styrofoam box has a surface area of 0.80 m 2 and a wall thickness of 2.0 cm. The temperature of the
inner surface is 5.0°C, and the outside temparature is 25°C. If it takes 8.0 hours for 5.0 kg of ice to melt in
the container, determine the thermal conductivity, k, of the Styrofoam.
Q/sec = k A DT/Dx => k = (Q/sec) Dx / (A DT)
In[34]:= BoxArea = .80; Dx = .020; DT = 25 - 5;
HeatToMeltIce = 5.0 * 3.33 * 10 5 ;
HeatFlowPerSecond = HeatToMeltIce ê H8 * 60 * 60L;
Print@"Rate of fheat flow = ", HeatFlowPerSecond, " Watts"D
kStyrofoam = HeatFlowPerSecond * Dx ê HBoxArea * DTL;
Print@"Thermal conductivity of Styrofoam = ", kStyrofoam, " Jês m °C"D
Rate of fheat flow = 57.8125 Watts
Thermal conductivity of Styrofoam = 0.0722656 Jês m °C
ü 47. A sphere that is a perfect blackbody radiator has a radius of 0.060m and is at 200°C in a room where
the temperature is 22°C. Calculate the net rate at which the sphere radiates energy. [In other words, the
difference between the radiation it gives off at 200°C and what it receives from the surroundings at 22°C.
And... remember to use Kelvins!]
In[75]:= s = 5.67 * 10 -8 ; Tradiator = 200 + 273.15; Troom = 20 + 273.15; sphereArea = 4 p .060 2 ;
NetHeatFlow = s * sphereArea * Tradiator 4 - s * sphereArea * Troom 4 ;
Print@"The net heat flowing out of the sphere is ", NetHeatFlow, " Watts."D
PrintA"The heat flowing out of the sphere is ", s * sphereArea * Tradiator 4 , " Watts."E
PrintA"The heat flowing back from the room is ", s * sphereArea * Troom 4 , " Watts."E
The net heat flowing out of the sphere is 109.612 Watts.
The heat flowing out of the sphere is 128.556 Watts.
The heat flowing back from the room is 18.9433 Watts.