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A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C
- a)1250°C
- b)825°C
- c)800°C
- d)885° C
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A copper ball of mass 100 gm is at a temperature T. It is dropped in a...
Heat given = Heat taken
(100)(0.1)(T–75)=(100)(0.1)(45)+(170)(1)(45)
10(T−75)=450+7650=8100
T−75=810
T=885degreeC
Most Upvoted Answer
A copper ball of mass 100 gm is at a temperature T. It is dropped in a...
°C. The specific heat capacity of copper is 0.39 J/g°C and the specific heat capacity of water is 4.18 J/g°C. Calculate the initial temperature of the copper ball.
To solve this problem, we can use the principle of conservation of energy. The heat lost by the copper ball is equal to the heat gained by the water and the calorimeter.
The heat lost by the copper ball can be calculated using the formula:
Q1 = m1 * c1 * (T1 - Tf)
Where:
Q1 = heat lost by the copper ball
m1 = mass of the copper ball
c1 = specific heat capacity of copper
T1 = initial temperature of the copper ball
Tf = final temperature of the system
Substituting the given values:
Q1 = 100 g * 0.39 J/g°C * (T1 - 75°C)
The heat gained by the water and the calorimeter can be calculated using the formula:
Q2 = (m2 + m3) * c2 * (Tf - T2)
Where:
Q2 = heat gained by the water and the calorimeter
m2 = mass of the water
m3 = mass of the calorimeter
c2 = specific heat capacity of water
T2 = initial temperature of the water and the calorimeter
Substituting the given values:
Q2 = (170 g + 100 g) * 4.18 J/g°C * (75°C - T2)
Since the heat lost by the copper ball is equal to the heat gained by the water and the calorimeter, we can set up the equation:
Q1 = Q2
100 g * 0.39 J/g°C * (T1 - 75°C) = (170 g + 100 g) * 4.18 J/g°C * (75°C - T2)
Simplifying the equation:
39(T1 - 75) = 270 * 4.18 * (75 - T2)
39T1 - 2925 = 270 * 4.18 * 75 - 270 * 4.18 * T2
39T1 - 2925 = 79065 - 1128.6T2
Rearranging the equation:
39T1 + 1128.6T2 = 79065 + 2925
39T1 + 1128.6T2 = 81990
Since we have two unknowns (T1 and T2), we need another equation to solve for both variables. We can use the fact that the final temperature of the system is 75°C:
Tf = (m1 * T1 + m2 * T2 + m3 * T2) / (m1 + m2 + m3)
Substituting the given values:
75°C = (100 g * T1 + 170 g * T2 + 100 g * T2) / (100 g + 170 g + 100 g)
75°C = (100T1 + 270T2 + 100T2) / 370 g
75°C = (100T1 + 370T2) / 370 g
Multiplying both sides of the equation by 370 g:
75°C * 370 g = 100T1 + 370T2
R
To solve this problem, we can use the principle of conservation of energy. The heat lost by the copper ball is equal to the heat gained by the water and the calorimeter.
The heat lost by the copper ball can be calculated using the formula:
Q1 = m1 * c1 * (T1 - Tf)
Where:
Q1 = heat lost by the copper ball
m1 = mass of the copper ball
c1 = specific heat capacity of copper
T1 = initial temperature of the copper ball
Tf = final temperature of the system
Substituting the given values:
Q1 = 100 g * 0.39 J/g°C * (T1 - 75°C)
The heat gained by the water and the calorimeter can be calculated using the formula:
Q2 = (m2 + m3) * c2 * (Tf - T2)
Where:
Q2 = heat gained by the water and the calorimeter
m2 = mass of the water
m3 = mass of the calorimeter
c2 = specific heat capacity of water
T2 = initial temperature of the water and the calorimeter
Substituting the given values:
Q2 = (170 g + 100 g) * 4.18 J/g°C * (75°C - T2)
Since the heat lost by the copper ball is equal to the heat gained by the water and the calorimeter, we can set up the equation:
Q1 = Q2
100 g * 0.39 J/g°C * (T1 - 75°C) = (170 g + 100 g) * 4.18 J/g°C * (75°C - T2)
Simplifying the equation:
39(T1 - 75) = 270 * 4.18 * (75 - T2)
39T1 - 2925 = 270 * 4.18 * 75 - 270 * 4.18 * T2
39T1 - 2925 = 79065 - 1128.6T2
Rearranging the equation:
39T1 + 1128.6T2 = 79065 + 2925
39T1 + 1128.6T2 = 81990
Since we have two unknowns (T1 and T2), we need another equation to solve for both variables. We can use the fact that the final temperature of the system is 75°C:
Tf = (m1 * T1 + m2 * T2 + m3 * T2) / (m1 + m2 + m3)
Substituting the given values:
75°C = (100 g * T1 + 170 g * T2 + 100 g * T2) / (100 g + 170 g + 100 g)
75°C = (100T1 + 270T2 + 100T2) / 370 g
75°C = (100T1 + 370T2) / 370 g
Multiplying both sides of the equation by 370 g:
75°C * 370 g = 100T1 + 370T2
R
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A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°Ca)1250°Cb)825°Cc)800°Cd)885° CCorrect answer is option 'D'. Can you explain this answer?
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A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°Ca)1250°Cb)825°Cc)800°Cd)885° CCorrect answer is option 'D'. 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 A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°Ca)1250°Cb)825°Cc)800°Cd)885° CCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°Ca)1250°Cb)825°Cc)800°Cd)885° CCorrect answer is option 'D'. Can you explain this answer?.
A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°Ca)1250°Cb)825°Cc)800°Cd)885° CCorrect answer is option 'D'. 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 A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°Ca)1250°Cb)825°Cc)800°Cd)885° CCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°Ca)1250°Cb)825°Cc)800°Cd)885° CCorrect answer is option 'D'. Can you explain this answer?.
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