Thermodymanics heat loss question. Please help

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  • #1
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Question: How much heat is lost in one hour through a 15 cm x 3.7m x 6.1m concrete floor if the inside temperature is 22 degrees c and the ground temperature is 13 degrees c?

some info:
thermal conductivity of concrete=1.1
change in temperature=8

Relative Equations:
kAchangeT/L

Attempt at problem:
1.1(.15x3.7)(22-13)/6.1
this is not giving me the right answer. the answer should be 5.363E6J
can someone explain to me what I am doing wrong.
thanks
 

Answers and Replies

  • #2
2,685
22
The change in temperature is 8 degrees celsius?

k = thermal conductivity
A = area
T = change in temerature
L = thickness of the surface

Is it more likely the concrete floor is 0.15mx3.7m with a thickness 6.1m or is it 3.7mx6.1m with a thickness of 0.15m?

You've got the values in the wrong place.
 
  • #3
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yes i tries rearranging the numbers several different ways but i still cant get the correct answer
 
  • #4
2,685
22
yes i tries rearranging the numbers several different ways but i still cant get the correct answer
22-13 isn't 8 for a start...

The equation only gives you the heat flow per second. You now need to convert it into heat flow per hour.
 
  • #5
394
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I suppose the answer should be
1.1(6.1x3.7)(22-13)/0.15*3600 but this is equal to 4,767E6
 
  • #6
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yes but even with that, io am getting the answer to br 3242.65 (still not correct)
 
  • #7
2,685
22
I suppose the answer should be
1.1(6.1x3.7)(22-13)/0.15*3600 but this is equal to 4,7676
I've worked it out and I get the exact answer specified by the OP.
 
  • #8
2,685
22
yes but even with that, io am getting the answer to br 3242.65 (still not correct)
The value you get should is per second, multiply by 3600 to get the value per hour.

3242.65 is incorrect.
 
  • #9
146
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but the answer should be 5.363 x 10^6 J
 
  • #10
2,685
22
k = 1.1
A = (3.7 x 6.1)
T = 9
L = 0.15

kAT/L = 1.1(3.7x6.1)9 / 0.15 = 223.443 / 0.15 = 1489.62

1489.62 x 3600 = ?
 
Last edited:
  • #11
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am i using the correct equation? could that be the problem?
 
  • #12
146
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Oh! thank you so much!
 
  • #13
2,685
22
am i using the correct equation? could that be the problem?
I've given the actual answer above.

I don't know what facenian did.

You need: Q/t = kAT/L

Where Q/t = heat loss per unit time.

When you solve the second part you get heat loss (Q) per second (t).

So you simply convert this to an hour.
 
  • #14
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Thank you(:
 
  • #15
394
15
I've worked it out and I get the exact answer specified by the OP.
you are right I used 22-13=8!
 

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