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Two plates A and B have thermal conductivities 84 Wm-1K-1 and 126 Wm-1K-1 respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of A and B are kept at 100 °C and 0 ºC respectively, then the temperature of the surface of contact in steady state is ________°C.
Correct answer is '40'. Can you explain this answer?
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Two plates A and B have thermal conductivities 84 Wm-1K-1and 126 Wm-1K...
Let the temperature of contact surface is T, then
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Two plates A and B have thermal conductivities 84 Wm-1K-1and 126 Wm-1K...
Understanding the Problem
Two plates A and B are in contact, with thermal conductivities of 84 Wm-1K-1 and 126 Wm-1K-1 respectively. The outer temperatures are maintained at 100 °C for plate A and 0 °C for plate B. We need to find the temperature at the interface in steady state.
Key Concepts
- Thermal Conductivity: A measure of a material's ability to conduct heat.
- Heat Transfer in Steady State: In steady state, the heat flow rate through both materials is equal.
Calculating Heat Flow
The heat flow (Q) through a material can be expressed as:
Q = (k * A * (T1 - T2)) / d
where:
- k = thermal conductivity
- A = surface area
- T1 and T2 = temperatures on either side
- d = thickness (same for both plates)
Setting Up the Equations
1. For Plate A:
- Q_A = (84 * A * (100 - T_contact)) / d
2. For Plate B:
- Q_B = (126 * A * (T_contact - 0)) / d
Since Q_A = Q_B in steady state:
84 * (100 - T_contact) = 126 * T_contact
Solving for T_contact
- Rearranging the equation:
84 * 100 - 84 * T_contact = 126 * T_contact
- Combine terms:
8400 = 210 * T_contact
- Therefore:
T_contact = 8400 / 210 = 40 °C
Conclusion
The temperature at the surface of contact in steady state is 40 °C. This result indicates that the temperature distribution is influenced by the thermal conductivities of the plates.
Two plates A and B are in contact, with thermal conductivities of 84 Wm-1K-1 and 126 Wm-1K-1 respectively. The outer temperatures are maintained at 100 °C for plate A and 0 °C for plate B. We need to find the temperature at the interface in steady state.
Key Concepts
- Thermal Conductivity: A measure of a material's ability to conduct heat.
- Heat Transfer in Steady State: In steady state, the heat flow rate through both materials is equal.
Calculating Heat Flow
The heat flow (Q) through a material can be expressed as:
Q = (k * A * (T1 - T2)) / d
where:
- k = thermal conductivity
- A = surface area
- T1 and T2 = temperatures on either side
- d = thickness (same for both plates)
Setting Up the Equations
1. For Plate A:
- Q_A = (84 * A * (100 - T_contact)) / d
2. For Plate B:
- Q_B = (126 * A * (T_contact - 0)) / d
Since Q_A = Q_B in steady state:
84 * (100 - T_contact) = 126 * T_contact
Solving for T_contact
- Rearranging the equation:
84 * 100 - 84 * T_contact = 126 * T_contact
- Combine terms:
8400 = 210 * T_contact
- Therefore:
T_contact = 8400 / 210 = 40 °C
Conclusion
The temperature at the surface of contact in steady state is 40 °C. This result indicates that the temperature distribution is influenced by the thermal conductivities of the plates.
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Two plates A and B have thermal conductivities 84 Wm-1K-1and 126 Wm-1K-1respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of A and B are kept at 100 °C and 0 ºC respectively, then the temperature of the surface of contact in steady state is ________°C.Correct answer is '40'. Can you explain this answer?
Question Description
Two plates A and B have thermal conductivities 84 Wm-1K-1and 126 Wm-1K-1respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of A and B are kept at 100 °C and 0 ºC respectively, then the temperature of the surface of contact in steady state is ________°C.Correct answer is '40'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two plates A and B have thermal conductivities 84 Wm-1K-1and 126 Wm-1K-1respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of A and B are kept at 100 °C and 0 ºC respectively, then the temperature of the surface of contact in steady state is ________°C.Correct answer is '40'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two plates A and B have thermal conductivities 84 Wm-1K-1and 126 Wm-1K-1respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of A and B are kept at 100 °C and 0 ºC respectively, then the temperature of the surface of contact in steady state is ________°C.Correct answer is '40'. Can you explain this answer?.
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