Ch. 13.3 Help
Page 1 of 1 • Share •
Ch. 13.3 Help
Question 1: (1 point)
A window pane with an 0.686 m2 area separates a room at 20 oC from the outside air at 10.41 oC. What is the power conducted through the 2.45 mm thick window pane (see sheet 22; use for the thermal conductivity of glass 0.8 W m1 oC1)? Indicate with a positive (negative) sign whether a styrofoam sheet of the same dimensions as the window pane conducts more (less) power.
How to solve:
Question 2: (1 point)
It takes 78.55 kcal to melt the amount of ice in a styrofoam box with a total wall area of 1.596 m2 and 1 cm thickness. How long does it take to melt the ice if the temperature outside the box is 16.1 oC (see sheet 22,24,25 use for the thermal conductivity of styrofoam 0.03 W m1 oC1)? Indicate with a positive (negative) sign whether you can (cannot) calculate the amount of ice present in the box.
How to solve:
Question 3: (1 point)
An object with an emissivity of 0.2749 and a surface area of 1.385 m2 radiates a power of 677.1 W. Assume that the emitted power by the surrounding of the object is negligible and calculate the temperature of the object in oC (see sheet 27,29).
How to solve:
Question 4: (1 point)
A naked person with a surface area of 1.785 m2 and a skin temperature of 29.94 oC rests in a room where the walls are at 0 oC. If the emissivity of the skin is assumed to be 1.1133 what is the net rate of energy loss of the person (heat loss) due to radiation (see sheet 26,27)? Indicate with a positive (negative) sign whether you can (cannot) neglect the power emitted by the walls and absorbed by the person for our typical 1% accuracy we work with.
How to solve:
A window pane with an 0.686 m2 area separates a room at 20 oC from the outside air at 10.41 oC. What is the power conducted through the 2.45 mm thick window pane (see sheet 22; use for the thermal conductivity of glass 0.8 W m1 oC1)? Indicate with a positive (negative) sign whether a styrofoam sheet of the same dimensions as the window pane conducts more (less) power.
How to solve:
Question 2: (1 point)
It takes 78.55 kcal to melt the amount of ice in a styrofoam box with a total wall area of 1.596 m2 and 1 cm thickness. How long does it take to melt the ice if the temperature outside the box is 16.1 oC (see sheet 22,24,25 use for the thermal conductivity of styrofoam 0.03 W m1 oC1)? Indicate with a positive (negative) sign whether you can (cannot) calculate the amount of ice present in the box.
How to solve:
Question 3: (1 point)
An object with an emissivity of 0.2749 and a surface area of 1.385 m2 radiates a power of 677.1 W. Assume that the emitted power by the surrounding of the object is negligible and calculate the temperature of the object in oC (see sheet 27,29).
How to solve:
Question 4: (1 point)
A naked person with a surface area of 1.785 m2 and a skin temperature of 29.94 oC rests in a room where the walls are at 0 oC. If the emissivity of the skin is assumed to be 1.1133 what is the net rate of energy loss of the person (heat loss) due to radiation (see sheet 26,27)? Indicate with a positive (negative) sign whether you can (cannot) neglect the power emitted by the walls and absorbed by the person for our typical 1% accuracy we work with.
How to solve:
sedwards Guest
#1,3,4
I have to run out hang in there and I will post the solutions to 1,3,and 4 around 8:45 tonight.
super mo Guest
#1,3,4
I got back early:
1: DeltaQ/Deltat= K*A*(T_2T_1)
A= 0.975m^2
T_1= 16.04deg C
T_2= 20deg C
K=0.8
L= 1.188mm = 0.001188m
0.8*0.975*((20 16.04)/0.001188) = 23662.6 answer neg.
3: T= ((P/A)/greek letter *e)^0.25
A= 1.1615m^2
P= 919
greek letter= 5.67*10^8
e= 0.4972
((919/1.615)/(5.67*10^8*0.4972))^0.25 = 376.927
answer to be in deg C so... 376.972273.15= 103.777 answer pos.
4:P= greek letter*e*A*(T^4T_s^4)
greek letter= 5.67*10^8
e= 0.8534
A= 1.763m^2
T= 49.91deg C= 323.06Kelvin
T_s= 0deg C= 273.15Kelvin
5.67*10^8*0.8534*1.763*((323.06^4)(273.15^4)) = 454.335 answer neg.
1: DeltaQ/Deltat= K*A*(T_2T_1)
A= 0.975m^2
T_1= 16.04deg C
T_2= 20deg C
K=0.8
L= 1.188mm = 0.001188m
0.8*0.975*((20 16.04)/0.001188) = 23662.6 answer neg.
3: T= ((P/A)/greek letter *e)^0.25
A= 1.1615m^2
P= 919
greek letter= 5.67*10^8
e= 0.4972
((919/1.615)/(5.67*10^8*0.4972))^0.25 = 376.927
answer to be in deg C so... 376.972273.15= 103.777 answer pos.
4:P= greek letter*e*A*(T^4T_s^4)
greek letter= 5.67*10^8
e= 0.8534
A= 1.763m^2
T= 49.91deg C= 323.06Kelvin
T_s= 0deg C= 273.15Kelvin
5.67*10^8*0.8534*1.763*((323.06^4)(273.15^4)) = 454.335 answer neg.
super mo Guest
#2
I have not checked with MapleTA as of yet but it is the correct answer for the CD.
A= 1.975
L= 1cm = 0.01m TR_1= 0deg C
T_2= 30.72dedg C
K=0.03
Q_melt= 71.39 kcal= 71390 cal*4.187 J= 298910
Part one:DeltaQ/Delta t= K*A*(T_2T_1)/L
0.03*1.975*(30.720)/0.01= 182.016
Part 2: t= Q_melt/(DeltaQ/Deltat)
298910/182.016= 1642.22 answer should be pos.
A= 1.975
L= 1cm = 0.01m TR_1= 0deg C
T_2= 30.72dedg C
K=0.03
Q_melt= 71.39 kcal= 71390 cal*4.187 J= 298910
Part one:DeltaQ/Delta t= K*A*(T_2T_1)/L
0.03*1.975*(30.720)/0.01= 182.016
Part 2: t= Q_melt/(DeltaQ/Deltat)
298910/182.016= 1642.22 answer should be pos.
super mo Guest
Re: Ch. 13.3 Help
I am still having problem with 3 and 4. MapleTA keeps saying that they are wrong.
cat Guest
Re: Ch. 13.3 Help
I hope this works. So this is a spreadsheet for 133. All you do is plug in the numbers and you get your answer. DON'T CONVERT!!!! Everything is done for you. I'm not sure if this is "legal" or not but I hope this helps you guys.
*Hope the link works *
*Hope the link works *
The Show Guest
Re: Ch. 13.3 Help
That's probably because the whole thing is a hoax, probably a virus or something.
Kathleen Guest
Re: Ch. 13.3 Help
Uhhno it's not a hoax. Maybe if you actually take out all the spaces then it'll work because I just tried it. And again, it's not a hoax. I help my friends with this so I don't have to go tell them how to compute the equations.
And if you think it's a virus, look at the extension. It's an Excel file (.xlsx).
BTW, number 3 on that sheet is wrong. Replace 522 with A4. The quiz is already due so I don't want to change it since nobody is going to use it anyways probably.
And if you think it's a virus, look at the extension. It's an Excel file (.xlsx).
BTW, number 3 on that sheet is wrong. Replace 522 with A4. The quiz is already due so I don't want to change it since nobody is going to use it anyways probably.
the show Guest
Re: Ch. 13.3 Help
"That's probably because the whole thing is a hoax, probably a virus or something."
haha how silly!
The spaces can be difficult to spot in the link. It might be better to use underscores next time. Try typing out the link by hand instead of removing the spaces. A+++++++ HIGHLY RECOMMENDED!!! WOULD DOWNLOAD FROM AGAIN!!!
haha how silly!
The spaces can be difficult to spot in the link. It might be better to use underscores next time. Try typing out the link by hand instead of removing the spaces. A+++++++ HIGHLY RECOMMENDED!!! WOULD DOWNLOAD FROM AGAIN!!!
Guest Guest
Re: Ch. 13.3 Help
ya no she's right i tried doing that link & that shIt totallyy messed up my mac
bree Guest
Page 1 of 1
Permissions in this forum:
You cannot reply to topics in this forum

