Heat loss per unit area per hour

  • Thread starter Amith2006
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  • #1
Amith2006
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Sir,
A man covers his body with a blanket of thickness 4 mm. His temperature is 37 degree Celsius and that of the atmosphere is 27 degree Celsius. If the coefficient of thermal conductivity of wool is 1.2 x 10^(-5) W/m.K , what is the heat lost per hour per square meter area?
I solved it in the following way:
Heat loss = KA(T1 – T2)t/y
Here K = 1.2 x 10^(-5) W/m.K
T1 – T2 = 10 K
y = 4 x 10^(-3)
Heat lost per unit area per hour = [1.2 x 10^(-5) x 10 x 3600]/(4 x 10^(-3))
= 108 Joule
Is it right?
 

Answers and Replies

  • #2
Andrew Mason
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Amith2006 said:
Sir,
A man covers his body with a blanket of thickness 4 mm. His temperature is 37 degree Celsius and that of the atmosphere is 27 degree Celsius. If the coefficient of thermal conductivity of wool is 1.2 x 10^(-5) W/m.K , what is the heat lost per hour per square meter area?
I solved it in the following way:
Heat loss = KA(T1 – T2)t/y
Here K = 1.2 x 10^(-5) W/m.K
T1 – T2 = 10 K
y = 4 x 10^(-3)
Heat lost per unit area per hour = [1.2 x 10^(-5) x 10 x 3600]/(4 x 10^(-3))
= 108 Joule
Is it right?
Your explanation of what you are doing and your units is not clear. If you are clear on the physics of what you doing, the correct answer will be much easier to see.

Heat conduction is given by:

[tex]dQ/dt = \frac{\kappa A(T_1 - T_2)}{d}[/tex]

where dQ/dt is the rate of heat transfer through the conducting material. You are looking for dQ/dt per unit area. So:

[tex]\frac{1}{A}\left(\frac{dQ}{dt}\right) = \frac{\kappa (T_1 - T_2)}{d}[/tex]

That will give you the answer in watts/m^2 or Joules/m^2 sec

You are looking for Joules/m^2hour = (Joules x 3600)/m^2(3600 sec).

So the answer is 108 Joules/m^2

AM
 
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