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What is the change in internal energy (in cal) when 0.004 kg. of air is heated from 0°C to 2°C. The specific heat of air at constant volume being 0.172 kilo cal/kg °C.
Correct answer is '1.376'. Can you explain this answer?
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What is the change in internal energy (in cal) when 0.004 kg. of air i...
From first law of thermodynamics
dQ = dU+ dW
where all quantities are measured in same units, say kilo cal.
Here air is heated at constant volume and hence no external work is done
i.e., dW = p. dV = 0.
Let m be the mass, cv the specific heat at constant volume and dT the rise in temperature of the air, the heat taken in by it is
dQ = m x cv x dT kilo cal.
Then from first law of thermodynamics
dU = dQ - dW = dQ
= m x cv x dT =0.004 x 0.172 x 2 = 1.376 x 10-3 kilo cal.
dQ = dU+ dW
where all quantities are measured in same units, say kilo cal.
Here air is heated at constant volume and hence no external work is done
i.e., dW = p. dV = 0.
Let m be the mass, cv the specific heat at constant volume and dT the rise in temperature of the air, the heat taken in by it is
dQ = m x cv x dT kilo cal.
Then from first law of thermodynamics
dU = dQ - dW = dQ
= m x cv x dT =0.004 x 0.172 x 2 = 1.376 x 10-3 kilo cal.
Most Upvoted Answer
What is the change in internal energy (in cal) when 0.004 kg. of air i...
The change in internal energy can be calculated using the equation:
ΔU = mcΔT
Where:
ΔU = change in internal energy
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
Given:
m = 0.004 kg
c (specific heat capacity of air) = 1 cal/g°C (constant pressure specific heat capacity)
ΔT = final temperature - initial temperature = Tfinal - Tinitial
The specific heat capacity of air is given in cal/g°C, so we need to convert the mass from kg to g:
m = 0.004 kg * 1000 g/kg = 4 g
Assuming the final temperature is Tfinal °C, and the initial temperature is Tinitial °C, the change in temperature is:
ΔT = Tfinal - Tinitial
Substituting the values into the equation:
ΔU = (4 g)(1 cal/g°C)(Tfinal - Tinitial)
Therefore, the change in internal energy in cal is (4 g)(1 cal/g°C)(Tfinal - Tinitial).
ΔU = mcΔT
Where:
ΔU = change in internal energy
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
Given:
m = 0.004 kg
c (specific heat capacity of air) = 1 cal/g°C (constant pressure specific heat capacity)
ΔT = final temperature - initial temperature = Tfinal - Tinitial
The specific heat capacity of air is given in cal/g°C, so we need to convert the mass from kg to g:
m = 0.004 kg * 1000 g/kg = 4 g
Assuming the final temperature is Tfinal °C, and the initial temperature is Tinitial °C, the change in temperature is:
ΔT = Tfinal - Tinitial
Substituting the values into the equation:
ΔU = (4 g)(1 cal/g°C)(Tfinal - Tinitial)
Therefore, the change in internal energy in cal is (4 g)(1 cal/g°C)(Tfinal - Tinitial).
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What is the change in internal energy (in cal) when 0.004 kg. of air is heated from 0°C to 2°C. The specific heat of air at constant volume being 0.172 kilo cal/kg °C.Correct answer is '1.376'. Can you explain this answer?
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What is the change in internal energy (in cal) when 0.004 kg. of air is heated from 0°C to 2°C. The specific heat of air at constant volume being 0.172 kilo cal/kg °C.Correct answer is '1.376'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about What is the change in internal energy (in cal) when 0.004 kg. of air is heated from 0°C to 2°C. The specific heat of air at constant volume being 0.172 kilo cal/kg °C.Correct answer is '1.376'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the change in internal energy (in cal) when 0.004 kg. of air is heated from 0°C to 2°C. The specific heat of air at constant volume being 0.172 kilo cal/kg °C.Correct answer is '1.376'. Can you explain this answer?.
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