Heat Transfer (Conduction) problem

In summary, To find the amount of thermal energy flowing through a solid oak wall per day, you can use the formula Q= k(A)(T2-T1)t/d, where Q is the thermal energy, k is the coefficient of thermal conductivity, A is the area of the wall, T2 and T1 are the temperatures on either side of the wall, t is the time in hours, and d is the thickness of the wall. In this specific problem, the area is 7.32 m^2 and the coefficient of thermal conductivity is 1.47 x 10^-1. Using the given temperatures of 21.1 C and -6.67 C, and a time of 24 hours, we can
  • #1
toddler
17
0

Homework Statement



Find the amount of thermal energy that flows per day through a solid oak wall 10.0 cm thick, 3.00 m long, and 2.44 m high, if the temp of the inside wall is 21.1 C while the temp of the outside wall is -6.67 C.

Homework Equations



Q= k(A)(T2-T1)t/d

coefficient of thermal conductivity for the oak is 1.47 x 10^-1


The Attempt at a Solution



This is a review question in my textbook and it shows:

=

(0.147 J/m s C) (24.0 m squared) (21.1 C - (-6.67 C))(24 hr)(3600 s/1 hr)
_______________________________________________________________
(0.100 m)




I understand the problem and the application, I am just unsure of how to get the area, A, for these problems. ...how did the 24 m^2 come about?
 
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  • #2
Hmm... the question says the area is 3.00 * 2.44 = 7.32 m^2.

I think the "24 m^2" is a typo.
 
  • #3



The 24 m^2 comes from the surface area of the oak wall. To calculate the surface area, you need to find the area of each side of the wall and then add them together. In this case, the wall is 10.0 cm thick, 3.00 m long, and 2.44 m high, so the surface area would be 2(3.00 m x 2.44 m) + 2(10.0 cm x 2.44 m) + 2(10.0 cm x 3.00 m) = 24.00 m^2. This is the total surface area of the wall, which is needed for the equation Q= k(A)(T2-T1)t/d. I hope this helps clarify the solution for you.
 

FAQ: Heat Transfer (Conduction) problem

What is heat transfer by conduction?

Heat transfer by conduction is the transfer of thermal energy between two objects that are in direct contact. This transfer occurs due to the collisions of particles within the objects, which causes energy to be transferred from the object with a higher temperature to the object with a lower temperature.

What factors affect heat transfer by conduction?

The rate of heat transfer by conduction can be affected by several factors, including the temperature difference between the objects, the material properties of the objects (such as thermal conductivity), the surface area in contact, and the distance between the objects.

How is heat transfer by conduction calculated?

The rate of heat transfer by conduction is calculated using the equation Q = (kA(T2-T1))/d, where Q is the heat transferred per unit time, k is the thermal conductivity of the material, A is the surface area in contact, T1 and T2 are the temperatures of the two objects, and d is the distance between the objects.

How does conduction differ from convection and radiation?

Conduction is the transfer of heat through direct contact between objects, while convection is the transfer of heat through the movement of fluids or gases. Radiation is the transfer of heat through electromagnetic waves. Unlike conduction and convection, radiation does not require a medium to transfer heat.

What are some real-world examples of heat transfer by conduction?

Some examples of heat transfer by conduction include using a metal spoon to stir hot soup (heat is transferred from the soup to the spoon), touching a hot pan (heat is transferred from the pan to your hand), and using a heating pad to warm up sore muscles (heat is transferred from the pad to your body).

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