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A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace is
- a)500°C
- b)400°C
- c)506.80 °C
- d)406.80 °C
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A piece of iron of mass 0.2 kg is kept inside a furnace, till it attai...
Heat lost by the piece of iron Q = M1C1 (θ1 – θ)
Here M1 = 0.2 Kg
C1 = 470 J Kg^-1 K^-1
Q = 0.2 X 470 (θ1 – 60) = 94 (θ1 – 60) ---- (1)
Heat gain by water and the calorimeter , Q = (M2 + w) C2 (θ - θ2)
M2 = 0.24 Kg
w = 0.01 Kg
θ2 = 20 deg C
C2 = Specific heat of water = 4200 J Kg^-1 K^-1
Q = (0.24 + 0.01) x 4200 X (60 – 20) = 42000 --(2)
From (1) and (2):
94 (θ1 – 60) = 42000
θ1 = 506.80 deg C
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Most Upvoted Answer
A piece of iron of mass 0.2 kg is kept inside a furnace, till it attai...
Use heat lost = heat gain . but the specific heat of iton must be given in the ques
Community Answer
A piece of iron of mass 0.2 kg is kept inside a furnace, till it attai...
°C. The final temperature of the mixture is found to be 30 °C. Determine the specific heat capacity of iron. Assume no heat is lost to the surroundings.
To solve this problem, we can use the principle of conservation of energy. The heat lost by the iron is equal to the heat gained by the water:
Heat lost by iron = Heat gained by water
The heat lost or gained can be calculated using the formula:
q = mcΔT
where q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
Let's calculate the heat lost by the iron:
q_iron = m_iron * c_iron * ΔT_iron
Given:
m_iron = 0.2 kg
ΔT_iron = (final temperature - initial temperature) = 30 - initial temperature
Now, let's calculate the heat gained by the water:
q_water = m_water * c_water * ΔT_water
Given:
m_water = 0.24 kg
c_water = specific heat capacity of water = 4,186 J/kg°C (approximately)
ΔT_water = (final temperature - initial temperature) = 30 - 20 = 10°C
Since the heat lost by the iron is equal to the heat gained by the water, we can set up the equation:
q_iron = q_water
m_iron * c_iron * ΔT_iron = m_water * c_water * ΔT_water
0.2 kg * c_iron * (30 - initial temperature) = 0.24 kg * 4,186 J/kg°C * 10°C
Now, we can solve for c_iron:
c_iron = (0.24 kg * 4,186 J/kg°C * 10°C) / (0.2 kg * (30 - initial temperature))
Let's assume the initial temperature of the iron is T_iron_initial. Substituting the values:
c_iron = (0.24 kg * 4,186 J/kg°C * 10°C) / (0.2 kg * (30 - T_iron_initial))
Now, we have all the values to calculate the specific heat capacity of iron.
To solve this problem, we can use the principle of conservation of energy. The heat lost by the iron is equal to the heat gained by the water:
Heat lost by iron = Heat gained by water
The heat lost or gained can be calculated using the formula:
q = mcΔT
where q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
Let's calculate the heat lost by the iron:
q_iron = m_iron * c_iron * ΔT_iron
Given:
m_iron = 0.2 kg
ΔT_iron = (final temperature - initial temperature) = 30 - initial temperature
Now, let's calculate the heat gained by the water:
q_water = m_water * c_water * ΔT_water
Given:
m_water = 0.24 kg
c_water = specific heat capacity of water = 4,186 J/kg°C (approximately)
ΔT_water = (final temperature - initial temperature) = 30 - 20 = 10°C
Since the heat lost by the iron is equal to the heat gained by the water, we can set up the equation:
q_iron = q_water
m_iron * c_iron * ΔT_iron = m_water * c_water * ΔT_water
0.2 kg * c_iron * (30 - initial temperature) = 0.24 kg * 4,186 J/kg°C * 10°C
Now, we can solve for c_iron:
c_iron = (0.24 kg * 4,186 J/kg°C * 10°C) / (0.2 kg * (30 - initial temperature))
Let's assume the initial temperature of the iron is T_iron_initial. Substituting the values:
c_iron = (0.24 kg * 4,186 J/kg°C * 10°C) / (0.2 kg * (30 - T_iron_initial))
Now, we have all the values to calculate the specific heat capacity of iron.
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A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °CCorrect answer is option 'C'. Can you explain this answer?
Question Description
A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °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 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °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 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °CCorrect answer is option 'C'. Can you explain this answer?.
A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °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 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °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 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °CCorrect answer is option 'C'. Can you explain this answer?.
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A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °CCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °CCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A piece of iron of mass 0.2 kg is kept inside a furnace, till it attains the temperature of the furnace. The hot piece of iron is then dropped into a calorimeter containing 0.24 kg of water at 20° C. The mixture attains an equilibrium temperature of 60°C. The temperature of the furnace isa)500°Cb)400°Cc)506.80 °Cd)406.80 °CCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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