Thermodynamics
- The internal energy change in a system that has absorbed 2 kcals of heat and done 500 J of work is :
-
View Hint View Answer Discuss in Forum
According to first law of thermodynamics Q = ∆U + W
∆U = Q – W
= 2 × 4.2 × 1000 – 500
= 8400 – 500 = 7900 JCorrect Option: C
According to first law of thermodynamics Q = ∆U + W
∆U = Q – W
= 2 × 4.2 × 1000 – 500
= 8400 – 500 = 7900 J
- An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram :
If Q1 , Q2 , Q3 indicate the heat a absorbed by the gas along the three processes and ∆U1 , ∆U2 , ∆U3 indicate the change in internal energy along the three processes respectively, then
-
View Hint View Answer Discuss in Forum
Initial and final condition is same for all process
∆U1 = ∆U2 = ∆U3
from first law of thermodynamics ∆Q = ∆U + ∆W
Work done ∆W1 > ∆W2 > ∆W3 (Area of P.V. graph)
So ∆Q1 > ∆Q2 > ∆Q3Correct Option: A
Initial and final condition is same for all process
∆U1 = ∆U2 = ∆U3
from first law of thermodynamics ∆Q = ∆U + ∆W
Work done ∆W1 > ∆W2 > ∆W3 (Area of P.V. graph)
So ∆Q1 > ∆Q2 > ∆Q3
- A system is taken from state a to state c by two paths adc and abc as shown in the figure. The internal energy at a is Ua = 10 J. Along the path adc the amount of heat absorbed δQ1 = 50 J and the work done δW1 = 20 J whereas along the path abc the heat absorbed δQ2 = 36 J. The amount of work done along the path abc is
-
View Hint View Answer Discuss in Forum
From first law of thermodynamics
Qadc = ∆Uadc + Wadc
50 J = ∆Uadc + 20 J
∆Uadc = 30 J
Again , Qabc = ∆Uabc + Wabc
Wabc = Qabc – ∆Uabc
= Qabc – ∆Uadc
= 36 J – 30 J = 6 JCorrect Option: A
From first law of thermodynamics
Qadc = ∆Uadc + Wadc
50 J = ∆Uadc + 20 J
∆Uadc = 30 J
Again , Qabc = ∆Uabc + Wabc
Wabc = Qabc – ∆Uabc
= Qabc – ∆Uadc
= 36 J – 30 J = 6 J
- The coefficient of performance of a refrigerator is 5. If the inside temperature of freezer is –20°C, then the temperature of the surroundings to which it rejects heat is
-
View Hint View Answer Discuss in Forum
Coefficient of performance, Cop = T2 T1 - T2 5 = 273 - 20 = 253 T1 - (273 - 20) T1 - 253
5T1 – (5 × 253) = 253
5T1 = 253 + (5 × 253) = 1518∴ T1 = 1518 = 303.6 5
or, T1 = 303.6 – 273 = 30.6 ≅ 31°C
Correct Option: D
Coefficient of performance, Cop = T2 T1 - T2 5 = 273 - 20 = 253 T1 - (273 - 20) T1 - 253
5T1 – (5 × 253) = 253
5T1 = 253 + (5 × 253) = 1518∴ T1 = 1518 = 303.6 5
or, T1 = 303.6 – 273 = 30.6 ≅ 31°C
- The efficiency of a Carnot engine operating between the temperatures of 100°C and –23°C will be
-
View Hint View Answer Discuss in Forum
η = 1 - T1 T2
T1 = –23°C = 250 K , T2 = 100°C = 373Kη = 1 - 250 = 373 - 250 373 373
Correct Option: C
η = 1 - T1 T2
T1 = –23°C = 250 K , T2 = 100°C = 373Kη = 1 - 250 = 373 - 250 373 373
