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A partition wall has two layers A and B in contact, each made of different material. They have the same thickness but the thermal conductivity of layer A is twice that of layer B. If the steady state temperature difference across the wall is 60 K, then the corresponding difference across the layer A is
- a)10 K
- b)20 K
- c)30 K
- d)40 K
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A partition wall has two layers A and B in contact, each made of diffe...
Explanation:
The rate of heat transfer through a material is given by:
Q = kA (ΔT / d)
Where Q is the rate of heat transfer, k is the thermal conductivity of the material, A is the area of the material, ΔT is the temperature difference across the material, and d is the thickness of the material.
In this case, we have a partition wall with two layers, A and B, in contact. Both layers have the same thickness, but the thermal conductivity of layer A is twice that of layer B. We are given that the steady state temperature difference across the wall is 60 K.
To find the corresponding temperature difference across layer A, we can use the equation above for each layer separately and then subtract the two equations. Let's call the temperature difference across layer A ΔTA and the temperature difference across layer B ΔTB.
For layer A:
Q = kA (ΔTA / d)
For layer B:
Q = kB A (ΔTB / d)
Since the two layers are in contact, the rate of heat transfer through them must be the same:
kA (ΔTA / d) = kB A (ΔTB / d)
Dividing both sides by A and rearranging:
ΔTA / ΔTB = kB / kA
We are given that kA = 2kB, so:
ΔTA / ΔTB = 1/2
ΔTB = 60 K (from the problem statement)
Solving for ΔTA:
ΔTA = (1/2) ΔTB = (1/2) (60 K) = 30 K
Therefore, the corresponding temperature difference across layer A is 30 K, which is answer choice (B).
The rate of heat transfer through a material is given by:
Q = kA (ΔT / d)
Where Q is the rate of heat transfer, k is the thermal conductivity of the material, A is the area of the material, ΔT is the temperature difference across the material, and d is the thickness of the material.
In this case, we have a partition wall with two layers, A and B, in contact. Both layers have the same thickness, but the thermal conductivity of layer A is twice that of layer B. We are given that the steady state temperature difference across the wall is 60 K.
To find the corresponding temperature difference across layer A, we can use the equation above for each layer separately and then subtract the two equations. Let's call the temperature difference across layer A ΔTA and the temperature difference across layer B ΔTB.
For layer A:
Q = kA (ΔTA / d)
For layer B:
Q = kB A (ΔTB / d)
Since the two layers are in contact, the rate of heat transfer through them must be the same:
kA (ΔTA / d) = kB A (ΔTB / d)
Dividing both sides by A and rearranging:
ΔTA / ΔTB = kB / kA
We are given that kA = 2kB, so:
ΔTA / ΔTB = 1/2
ΔTB = 60 K (from the problem statement)
Solving for ΔTA:
ΔTA = (1/2) ΔTB = (1/2) (60 K) = 30 K
Therefore, the corresponding temperature difference across layer A is 30 K, which is answer choice (B).
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Community Answer
A partition wall has two layers A and B in contact, each made of diffe...
Heat=KA∆t
2KA∆t1 =KA∆t2
2/1=∆t2/∆t1
2/1 =60- ∆t1/∆t1
∆t1=20K
2KA∆t1 =KA∆t2
2/1=∆t2/∆t1
2/1 =60- ∆t1/∆t1
∆t1=20K
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A partition wall has two layers A and B in contact, each made of different material. They have the same thickness but the thermal conductivity of layer A is twice that of layer B. If the steady state temperature difference across the wall is 60 K, then the corresponding difference across the layer A isa)10 Kb)20 Kc)30 Kd)40 KCorrect answer is option 'B'. Can you explain this answer?
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