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A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.
- a)500°C
- b)900°C
- c)953.6∘C
- d)706.80 °C
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A piece of iron of mass 100g is kept inside a furnace for a long time ...
Mass of Iron = 100g
Water Eq of caloriemeter = 10g
Mass of water = 240g
Let the Temp. of surface = 0ºC
Siron = 470J/kg°C
Water Eq of caloriemeter = 10g
Mass of water = 240g
Let the Temp. of surface = 0ºC
Siron = 470J/kg°C
Total heat gained = Total heat lost.
So,100/1000× 470 × (θ – 60) = 250/1000 × 4200 × (60 – 20)
⇒ 47θ – 47 × 60 = 25 × 42 × 40
⇒ θ = 4200 + 2820/47= 44820/47 =953.61°C
⇒ 47θ – 47 × 60 = 25 × 42 × 40
⇒ θ = 4200 + 2820/47= 44820/47 =953.61°C
Most Upvoted Answer
A piece of iron of mass 100g is kept inside a furnace for a long time ...
Mass of iron = 100 g
Water equivalent of calorimeter = 10 g
Mass of water = 240 g
Let the temperature of the surface be ToC
Specific heat of iron = S = 470Jkg−1C−1
Total heat gained = Total heat lost
1000100×470×(T−60)=1000250(4200)(60−20)
T=953.610C
Yoyo
Water equivalent of calorimeter = 10 g
Mass of water = 240 g
Let the temperature of the surface be ToC
Specific heat of iron = S = 470Jkg−1C−1
Total heat gained = Total heat lost
1000100×470×(T−60)=1000250(4200)(60−20)
T=953.610C
Yoyo
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Community Answer
A piece of iron of mass 100g is kept inside a furnace for a long time ...
°C. The initial temperature of the iron is also 20°C. After some time, the final temperature of the system is reached at 50°C. Calculate the specific heat capacity of iron.
To solve this problem, we can use the principle of conservation of energy. The heat gained by the water and the calorimeter is equal to the heat lost by the iron.
The heat gained by the water and the calorimeter can be calculated using the formula:
Q = m * c * ΔT
where:
Q is the heat gained by the water and the calorimeter
m is the mass of the water and the calorimeter (240g + 10g = 250g)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (50°C - 20°C = 30°C)
Q = 250g * 4.18 J/g°C * 30°C
Q = 31275 J
The heat lost by the iron can be calculated using the formula:
Q = m * c * ΔT
where:
Q is the heat lost by the iron
m is the mass of the iron (100g)
c is the specific heat capacity of iron (unknown)
ΔT is the change in temperature (50°C - 20°C = 30°C)
Q = 100g * c * 30°C
Q = 3000c J
Since the heat gained by the water and the calorimeter is equal to the heat lost by the iron, we can set up the equation:
31275 J = 3000c J
Simplifying the equation:
c = 31275 J / 3000 J/g
c = 10.425 J/g°C
Therefore, the specific heat capacity of iron is 10.425 J/g°C.
To solve this problem, we can use the principle of conservation of energy. The heat gained by the water and the calorimeter is equal to the heat lost by the iron.
The heat gained by the water and the calorimeter can be calculated using the formula:
Q = m * c * ΔT
where:
Q is the heat gained by the water and the calorimeter
m is the mass of the water and the calorimeter (240g + 10g = 250g)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (50°C - 20°C = 30°C)
Q = 250g * 4.18 J/g°C * 30°C
Q = 31275 J
The heat lost by the iron can be calculated using the formula:
Q = m * c * ΔT
where:
Q is the heat lost by the iron
m is the mass of the iron (100g)
c is the specific heat capacity of iron (unknown)
ΔT is the change in temperature (50°C - 20°C = 30°C)
Q = 100g * c * 30°C
Q = 3000c J
Since the heat gained by the water and the calorimeter is equal to the heat lost by the iron, we can set up the equation:
31275 J = 3000c J
Simplifying the equation:
c = 31275 J / 3000 J/g
c = 10.425 J/g°C
Therefore, the specific heat capacity of iron is 10.425 J/g°C.
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A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. Can you explain this answer?
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A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. 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 A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. 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 A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. Can you explain this answer?.
A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. 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 A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. 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 A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. Can you explain this answer?.
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A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A piece of iron of mass 100g is kept inside a furnace for a long time and Jthen put in a calorimeter of water equivalent 10g containing 240g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470J/kg-°C.a)500°Cb)900°Cc)953.6Cd)706.80 °CCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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