Heat loss per unit area per hour

[Amith2006]

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?

 

[Andrew Mason]

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

 

[kyle8921]

Yes, your calculation is correct. The formula you used, Q = KA(T1-T2)t/y, is the correct equation for calculating heat loss per unit area per hour. Plugging in the given values, you correctly calculated the heat loss to be 108 Joule. Good job!