Two identical bars are conducting heat from a region of higher temperature to one of lower temperature. In arrangement A, the bars conduct 80 J of heat in a certain amount of time. How much heat is conducted in B during the same time

Q' = 320 J

Explanation:

The arrangements are given in attachment. The Fourier's Law of Heat Conduction states that:

Q = KAΔT/L

where,

Q = Heat Transferred

K = Constant (Conduction Coefficient)

A = Surface Area of Heat Transfer

ΔT = Difference of Temperature between two surfaces

L = Length between surfaces

For Arrangement AL

Q = 80 J

Therefore,

80 = KAΔT/L   ------------- equation (1)

Now, for arrangement B:

A' = 2 A (As, the rods are now connected in parallel with each other)

L' = L/2

Therefore,

Q' = K(2A)ΔT/(L/2)

Q' = 4 KAΔT/L

using equation (1)

Q' = 4(80 J)

Q' = 320 J

answer; ///i believe that the correct answer is (45*) at a 45 degree check first good

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