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A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?
- a)26.8 cm
- b)25.8 cm
- c)24.8 cm
- d)23.8 cm
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A thermos flask has a double walled bottle and the space between the w...
Q = k A (t 1 – t 2)/δ. So, δ = 0.268 m = 26.8 cm.
Most Upvoted Answer
A thermos flask has a double walled bottle and the space between the w...
To determine the thickness of cork required to achieve the same insulating effect as the evacuated double-walled thermos flask, we can use the concept of thermal resistance.
1. Define the thermal resistance of the flask:
The thermal resistance of the double-walled flask is given by the equation:
R_flask = (1 / (2 * π * k * L)) * ln(r2 / r1)
Where:
- k is the thermal conductivity of the silver-plated walls (unknown)
- L is the thickness of the walls (unknown)
- r2 is the outer radius of the flask
- r1 is the inner radius of the flask
2. Define the thermal resistance of the cork:
The thermal resistance of the cork is given by the equation:
R_cork = (L_cork / (k_cork * A))
Where:
- L_cork is the thickness of the cork (unknown)
- k_cork is the thermal conductivity of the cork (0.03 W/m degree)
- A is the surface area of the flask
3. Equate the two thermal resistances:
R_flask = R_cork
4. Solve for the thickness of the cork:
(L / (2 * π * k * L)) * ln(r2 / r1) = (L_cork / (k_cork * A))
5. Simplify the equation:
(1 / (2 * π * k)) * ln(r2 / r1) = (L_cork / (k_cork * A))
6. Substitute the known values:
- π = 3.1416
- k = unknown
- r2 = unknown
- r1 = unknown
- k_cork = 0.03 W/m degree
- A = unknown
7. Rearrange the equation to solve for k:
k = (1 / (2 * π)) * (ln(r2 / r1) / (L_cork / (k_cork * A)))
8. Substitute the known values:
k = (1 / (2 * π)) * (ln(r2 / r1) / (0.03 * A))
9. Simplify the equation:
k = (33.333 * ln(r2 / r1)) / A
10. Calculate the thermal conductivity of the cork:
k = (33.333 * ln(r2 / r1)) / A
11. Calculate the thickness of the cork:
L_cork = (k * A * R_flask) / (33.333 * ln(r2 / r1))
12. Substitute the known values:
- k = 0.03 W/m degree
- A = unknown
- R_flask = (1 / (2 * π * k * L)) * ln(r2 / r1)
- r2 = unknown
- r1 = unknown
13. Rearrange the equation to solve for A:
A = (L_cork * 33.333 * ln(r2 / r1)) / (k * R_flask)
14. Substitute the known values:
A = (L_cork * 33.333 * ln(r2 / r1)) / (0.03 * R_flask)
15. Calculate the surface area of the flask:
A = (L_cork * 1111.1 * ln(r2 / r1)) / R_flask
16. The thickness of
1. Define the thermal resistance of the flask:
The thermal resistance of the double-walled flask is given by the equation:
R_flask = (1 / (2 * π * k * L)) * ln(r2 / r1)
Where:
- k is the thermal conductivity of the silver-plated walls (unknown)
- L is the thickness of the walls (unknown)
- r2 is the outer radius of the flask
- r1 is the inner radius of the flask
2. Define the thermal resistance of the cork:
The thermal resistance of the cork is given by the equation:
R_cork = (L_cork / (k_cork * A))
Where:
- L_cork is the thickness of the cork (unknown)
- k_cork is the thermal conductivity of the cork (0.03 W/m degree)
- A is the surface area of the flask
3. Equate the two thermal resistances:
R_flask = R_cork
4. Solve for the thickness of the cork:
(L / (2 * π * k * L)) * ln(r2 / r1) = (L_cork / (k_cork * A))
5. Simplify the equation:
(1 / (2 * π * k)) * ln(r2 / r1) = (L_cork / (k_cork * A))
6. Substitute the known values:
- π = 3.1416
- k = unknown
- r2 = unknown
- r1 = unknown
- k_cork = 0.03 W/m degree
- A = unknown
7. Rearrange the equation to solve for k:
k = (1 / (2 * π)) * (ln(r2 / r1) / (L_cork / (k_cork * A)))
8. Substitute the known values:
k = (1 / (2 * π)) * (ln(r2 / r1) / (0.03 * A))
9. Simplify the equation:
k = (33.333 * ln(r2 / r1)) / A
10. Calculate the thermal conductivity of the cork:
k = (33.333 * ln(r2 / r1)) / A
11. Calculate the thickness of the cork:
L_cork = (k * A * R_flask) / (33.333 * ln(r2 / r1))
12. Substitute the known values:
- k = 0.03 W/m degree
- A = unknown
- R_flask = (1 / (2 * π * k * L)) * ln(r2 / r1)
- r2 = unknown
- r1 = unknown
13. Rearrange the equation to solve for A:
A = (L_cork * 33.333 * ln(r2 / r1)) / (k * R_flask)
14. Substitute the known values:
A = (L_cork * 33.333 * ln(r2 / r1)) / (0.03 * R_flask)
15. Calculate the surface area of the flask:
A = (L_cork * 1111.1 * ln(r2 / r1)) / R_flask
16. The thickness of
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A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?a)26.8 cmb)25.8 cmc)24.8 cmd)23.8 cmCorrect answer is option 'A'. Can you explain this answer?
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A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?a)26.8 cmb)25.8 cmc)24.8 cmd)23.8 cmCorrect answer is option 'A'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?a)26.8 cmb)25.8 cmc)24.8 cmd)23.8 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?a)26.8 cmb)25.8 cmc)24.8 cmd)23.8 cmCorrect answer is option 'A'. Can you explain this answer?.
A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?a)26.8 cmb)25.8 cmc)24.8 cmd)23.8 cmCorrect answer is option 'A'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?a)26.8 cmb)25.8 cmc)24.8 cmd)23.8 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?a)26.8 cmb)25.8 cmc)24.8 cmd)23.8 cmCorrect answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?a)26.8 cmb)25.8 cmc)24.8 cmd)23.8 cmCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Chemical Engineering tests.
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