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In the temperature range 100-1000 C, the molar specific heat of a metal varies with temperature T (measured in degrees Celsius) according to the formula Cp=(1 + T/5) J - deg C-1 mol-1. If 0.2 kg of the metal at 600c is brought in thermal contact with 0.1 kg of the same metal at 300 c, the final equilibrium temperature, in deg c, will be ________ .
Correct answer is '519'. Can you explain this answer?
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In the temperature range 100-1000 C, the molar specific heat of a meta...
We know that in the system
Heat gain by the system = Heat released by the system
Heat gain by the system = Heat released by the system
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In the temperature range 100-1000 C, the molar specific heat of a meta...
Given information:
- Temperature range: 100-1000 °C
- Molar specific heat of the metal: Cp = (1 - T/5) J - °C⁻¹ mol⁻¹
- Mass of metal 1: 0.2 kg
- Temperature of metal 1: 600 °C
- Mass of metal 2: 0.1 kg
- Temperature of metal 2: 300 °C
To find:
The final equilibrium temperature when metal 1 and metal 2 are brought in thermal contact.
Solution:
Step 1: Calculate the heat transferred from metal 1 to metal 2.
- Heat transferred = mass × specific heat × change in temperature
- Heat transferred from metal 1 to metal 2 = (0.2 kg) × Cp × (600 °C - final temperature)
Step 2: Calculate the heat transferred from metal 2 to metal 1.
- Heat transferred = mass × specific heat × change in temperature
- Heat transferred from metal 2 to metal 1 = (0.1 kg) × Cp × (final temperature - 300 °C)
Step 3: Set the two heat transfers equal to each other and solve for the final temperature.
- (0.2 kg) × Cp × (600 °C - final temperature) = (0.1 kg) × Cp × (final temperature - 300 °C)
Step 4: Simplify the equation and solve for the final temperature.
- 0.2 × (1 - (600/5)) × (600 - final temperature) = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
- 0.2 × (1 - 120) × (600 - final temperature) = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
- -0.2 × (600 - final temperature) = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
Step 5: Solve the equation to find the final temperature.
- -120 + 0.2 × final temperature = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.1 × (1 - (final temperature - 300)/5) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.1 × ((5 - (final temperature - 300))/5) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.02 × (5 - (final temperature - 300)) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.02 × (5 - final temperature + 300) × (final temperature - 300)
- Temperature range: 100-1000 °C
- Molar specific heat of the metal: Cp = (1 - T/5) J - °C⁻¹ mol⁻¹
- Mass of metal 1: 0.2 kg
- Temperature of metal 1: 600 °C
- Mass of metal 2: 0.1 kg
- Temperature of metal 2: 300 °C
To find:
The final equilibrium temperature when metal 1 and metal 2 are brought in thermal contact.
Solution:
Step 1: Calculate the heat transferred from metal 1 to metal 2.
- Heat transferred = mass × specific heat × change in temperature
- Heat transferred from metal 1 to metal 2 = (0.2 kg) × Cp × (600 °C - final temperature)
Step 2: Calculate the heat transferred from metal 2 to metal 1.
- Heat transferred = mass × specific heat × change in temperature
- Heat transferred from metal 2 to metal 1 = (0.1 kg) × Cp × (final temperature - 300 °C)
Step 3: Set the two heat transfers equal to each other and solve for the final temperature.
- (0.2 kg) × Cp × (600 °C - final temperature) = (0.1 kg) × Cp × (final temperature - 300 °C)
Step 4: Simplify the equation and solve for the final temperature.
- 0.2 × (1 - (600/5)) × (600 - final temperature) = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
- 0.2 × (1 - 120) × (600 - final temperature) = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
- -0.2 × (600 - final temperature) = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
Step 5: Solve the equation to find the final temperature.
- -120 + 0.2 × final temperature = 0.1 × (1 - ((final temperature - 300)/5)) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.1 × (1 - (final temperature - 300)/5) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.1 × ((5 - (final temperature - 300))/5) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.02 × (5 - (final temperature - 300)) × (final temperature - 300)
- -120 + 0.2 × final temperature = 0.02 × (5 - final temperature + 300) × (final temperature - 300)
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In the temperature range 100-1000 C, the molar specific heat of a meta...
We know that in the system
Heat gain by the system = Heat released by the system
Heat gain by the system = Heat released by the system
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In the temperature range 100-1000 C, the molar specific heat of a metal varieswith temperature T (measured in degrees Celsius) according to the formulaCp=(1 + T/5) J - deg C-1 mol-1. If 0.2 kg of the metal at 600c is brought in thermalcontact with 0.1 kg of the same metal at 300 c, the final equilibrium temperature, in deg c, will be ________ .Correct answer is '519'. Can you explain this answer?
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In the temperature range 100-1000 C, the molar specific heat of a metal varieswith temperature T (measured in degrees Celsius) according to the formulaCp=(1 + T/5) J - deg C-1 mol-1. If 0.2 kg of the metal at 600c is brought in thermalcontact with 0.1 kg of the same metal at 300 c, the final equilibrium temperature, in deg c, will be ________ .Correct answer is '519'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about In the temperature range 100-1000 C, the molar specific heat of a metal varieswith temperature T (measured in degrees Celsius) according to the formulaCp=(1 + T/5) J - deg C-1 mol-1. If 0.2 kg of the metal at 600c is brought in thermalcontact with 0.1 kg of the same metal at 300 c, the final equilibrium temperature, in deg c, will be ________ .Correct answer is '519'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the temperature range 100-1000 C, the molar specific heat of a metal varieswith temperature T (measured in degrees Celsius) according to the formulaCp=(1 + T/5) J - deg C-1 mol-1. If 0.2 kg of the metal at 600c is brought in thermalcontact with 0.1 kg of the same metal at 300 c, the final equilibrium temperature, in deg c, will be ________ .Correct answer is '519'. Can you explain this answer?.
Solutions for In the temperature range 100-1000 C, the molar specific heat of a metal varieswith temperature T (measured in degrees Celsius) according to the formulaCp=(1 + T/5) J - deg C-1 mol-1. If 0.2 kg of the metal at 600c is brought in thermalcontact with 0.1 kg of the same metal at 300 c, the final equilibrium temperature, in deg c, will be ________ .Correct answer is '519'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
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