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Direction (Q. Nos. 12 - 14) This section contains 3 questions. when worked out will result in an integer from 0 to 9 (both inclusive).
Q. An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?
Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.
Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.
Correct answer is '2'. Can you explain this answer?
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Direction (Q. Nos. 12- 14) This section contains 3questions. when work...
Heat lost by iron = Heat gained by water
Let the equilibrium temperature be T.
mIronsIron (352-T) = mwaterswater(T-292)
88.5×0.449×(352-T) = 244×4.18×(T-292)
On calculation approximation,
40(352-T) = 1020(T-292)
0.04(352-T) = (T-292)
T = 306/1.04 = 294 K
So, change in temperature of water = 294-292 = 2K
Let the equilibrium temperature be T.
mIronsIron (352-T) = mwaterswater(T-292)
88.5×0.449×(352-T) = 244×4.18×(T-292)
On calculation approximation,
40(352-T) = 1020(T-292)
0.04(352-T) = (T-292)
T = 306/1.04 = 294 K
So, change in temperature of water = 294-292 = 2K
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Direction (Q. Nos. 12- 14) This section contains 3questions. when work...
Given:
- Mass of iron piece = 88.5 g
- Initial temperature of iron = 352.0 K
- Mass of water in beaker = 244 g
- Initial temperature of water = 292.0 K
- Specific heat of water = 4.184 J/g K
- Specific heat of iron = 0.449 J/g K
To find:
Change in the temperature of water when thermal equilibrium is reached.
Solution:
Step 1: Calculate the heat absorbed by the iron piece.
- The heat absorbed by the iron can be calculated using the formula: Q = mcΔT, where Q is the heat absorbed, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.
- Let's calculate the heat absorbed by the iron:
Q_iron = (mass of iron) x (specific heat of iron) x (change in temperature of iron)
Q_iron = 88.5 g x 0.449 J/g K x (Final temperature of iron - Initial temperature of iron)
Since the final temperature of iron is not given, we cannot calculate the exact value of Q_iron.
Step 2: Calculate the heat released by the water.
- The heat released by the water can also be calculated using the formula: Q = mcΔT.
- Let's calculate the heat released by the water:
Q_water = (mass of water) x (specific heat of water) x (change in temperature of water)
Q_water = 244 g x 4.184 J/g K x (Final temperature of water - Initial temperature of water)
Since the final temperature of water is not given, we cannot calculate the exact value of Q_water.
Step 3: Use the principle of energy conservation.
- In thermal equilibrium, the heat absorbed by the iron is equal to the heat released by the water.
- Therefore, we can equate Q_iron and Q_water:
Q_iron = Q_water
88.5 g x 0.449 J/g K x (Final temperature of iron - Initial temperature of iron) = 244 g x 4.184 J/g K x (Final temperature of water - Initial temperature of water)
Step 4: Solve the equation to find the change in temperature of water.
- Rearrange the equation to solve for the change in temperature of water:
(Final temperature of water - Initial temperature of water) = (88.5 g x 0.449 J/g K x (Final temperature of iron - Initial temperature of iron)) / (244 g x 4.184 J/g K)
Step 5: Calculate the change in temperature of water.
- Substitute the given values into the equation:
(Final temperature of water - 292.0 K) = (88.5 g x 0.449 J/g K x (Final temperature of iron - 352.0 K)) / (244 g x 4.184 J/g K)
Simplifying the equation:
(Final temperature of water - 292.0 K) = (0.449/4.184) x (88.5 g / 244 g) x (Final temperature of iron - 352.0 K)
(Final temperature of
- Mass of iron piece = 88.5 g
- Initial temperature of iron = 352.0 K
- Mass of water in beaker = 244 g
- Initial temperature of water = 292.0 K
- Specific heat of water = 4.184 J/g K
- Specific heat of iron = 0.449 J/g K
To find:
Change in the temperature of water when thermal equilibrium is reached.
Solution:
Step 1: Calculate the heat absorbed by the iron piece.
- The heat absorbed by the iron can be calculated using the formula: Q = mcΔT, where Q is the heat absorbed, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.
- Let's calculate the heat absorbed by the iron:
Q_iron = (mass of iron) x (specific heat of iron) x (change in temperature of iron)
Q_iron = 88.5 g x 0.449 J/g K x (Final temperature of iron - Initial temperature of iron)
Since the final temperature of iron is not given, we cannot calculate the exact value of Q_iron.
Step 2: Calculate the heat released by the water.
- The heat released by the water can also be calculated using the formula: Q = mcΔT.
- Let's calculate the heat released by the water:
Q_water = (mass of water) x (specific heat of water) x (change in temperature of water)
Q_water = 244 g x 4.184 J/g K x (Final temperature of water - Initial temperature of water)
Since the final temperature of water is not given, we cannot calculate the exact value of Q_water.
Step 3: Use the principle of energy conservation.
- In thermal equilibrium, the heat absorbed by the iron is equal to the heat released by the water.
- Therefore, we can equate Q_iron and Q_water:
Q_iron = Q_water
88.5 g x 0.449 J/g K x (Final temperature of iron - Initial temperature of iron) = 244 g x 4.184 J/g K x (Final temperature of water - Initial temperature of water)
Step 4: Solve the equation to find the change in temperature of water.
- Rearrange the equation to solve for the change in temperature of water:
(Final temperature of water - Initial temperature of water) = (88.5 g x 0.449 J/g K x (Final temperature of iron - Initial temperature of iron)) / (244 g x 4.184 J/g K)
Step 5: Calculate the change in temperature of water.
- Substitute the given values into the equation:
(Final temperature of water - 292.0 K) = (88.5 g x 0.449 J/g K x (Final temperature of iron - 352.0 K)) / (244 g x 4.184 J/g K)
Simplifying the equation:
(Final temperature of water - 292.0 K) = (0.449/4.184) x (88.5 g / 244 g) x (Final temperature of iron - 352.0 K)
(Final temperature of
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Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer?
Question Description
Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer?.
Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer?.
Solutions for Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer?, a detailed solution for Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer? has been provided alongside types of Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Direction (Q. Nos. 12- 14) This section contains 3questions. when worked out will result in an integer from 0 to 9 (both inclusive).Q.An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?Specific heat of water and iron are (4.184 Jg-1 K-1) and (0.449 Jg-1 K-1), respectively.Correct answer is '2'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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