Ask a Teacher
CLASS NOTE OF THERMODYNAMIC WITH MORE PROBLEMS(QUESTIONS) |
P1 = 1 atm = 1.013 × 105 Nm-2, V1/V2 = 2
= 1.013 × 105 × 2.64 = 2.694 × 105 Nm-2
Here P1 = 6 atm, T1 = 15 + 273 = 288 K, γ = 1.4 P2 = 1 atm
Taking
log both sides, we get = 1.4(2.4742) - 0.4 (0.7782) = 3.1526 or
= 178.6 - 273 =
94.4oC.
Here T1 = 27 + 273 = 300 K, P1 = 1 atm, V2 = V1/2, γ = 1.42 (i) As P2V2γ = P1V1γ
or
T2
= 401.3 - 273 = 128.3oC.
Here T1 = 15 + 273 = 288 K, P1/P2 = 10 For adiabatic process,
Taking log of both sides 1.41 (log 288 - log T2) = 0.41 log 10 = 0.41 × 1
T2 = Antilog (2.1701) = 147.9 K or -125oC 5. A Carnot engine operates between 227oC and 127oC. If it absorbs 60 × 104 calorie at higher temperature, how much work per cycle can the engine perform?
Here T1 = 227 + 273 = 500 K, T2 = 127 + 273 = 400K As
or W = 12 × 104 cal
= 12 × 104
× 4.2 J = 5.04 × 105
J.
6.How much energy in watt hour may be required to convert 2 kg of water into ice at 0oC, assuming that the refrigeration is ideal? Given temperature of freezer is -15oC, room temperature is 25oC and initial temperature of water is 25oC.
Here T1 = 25 + 273 = 298 K, T2 = -15 + 273 = 258 K Specific heat of water, c = 4.2 × 103 J kg-1 Latent heat of ice, L = 3.36 × 105 J kg-1 Amount of heat required to be removed from 2 kg of water at 25oC to change it into ice at 0oC, Q2 = Mc(θ2 - θ1) + ML = 2 × 4.2 × 103 (25 - 0) + 2 × 3.36 × 105 = 2.1 × 105 + 6.72 × 105 = 8.82 × 105J Heat rejected to the surroundings,
Energy supplied to convert water into ice, W = Q1 - Q2 = (10.15 - 8.82) × 105 = 1.33 × 105 J =
7. Assuming that a domestic refrigerator can be regarded as a reversible engine working between the temperature of melting ice and that of the atmosphere (17oC), calculate the energy which must be supplied to freeze one kilogram of water already at 0oC. Here T2 = 273 K, T1 = 17 + 273 = 290 K Heat required to be removed to freeze 1 kg of water at 0oC Q2 = mL = 1 × 80,000 cal = 80,000 × 4.2 J
8.A refrigerator whose coefficient of performance is 5 extracts heat from the cooling compartment at the rate of 250 J/cycle?How much electric energy is spent per cycle? How much heat per cycle is discharged to the room?
Here Q2 = 250J/cycle, β = 5
Heat discharged to the room per cycle, Q1
= Q2
+ W = 250 + 50 = 300 J.
9. A refigerator has to transfer an average of 263 J of heat per second from temperature -10oC to 25oC. Calculate the average power consumed, assuming no energy losses in the process.
Here T1 = 25 + 273 = 298 K, T2 = 10 + 273 = 263 K, Q2 = 263 J
Average power consumed, W = Q1 - Q2 = 298 - 263 = 35 Js-1 = 35 W.
10.An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q1 = 5960 J, Q2 = -5585J, Q3 = -2980 J and Q4 = 3645 J respectively. The corresponding works involved are W1 = 2200 J, W2 = - 825 J, W3 = -1100 J and W4 respectively. Find (i) W4 and (ii) efficiency of the cycle.
(i) By the first law of thermodynamics, dQ = dU + dW But for a cyclic process, dU = 0 ∴ dQ = dW or Q1 + Q2 + Q3 + Q4 = W1 + W2 + W3 + W4 or 5960 - 5585 - 2980 + 3645 = 2200 - 825 - 1100 + W4 or W4 = (5960 + 3645 + 825 + 1100) - (5585 + 2980 + 2200) = 11530 - 10765 = 765 J. (ii) Efficiency,
11. Let the temperatures T1 and T2 of the two heat reservoirs in an ideal carnot engine be 1500oC and 500oC respectively. Which of these, increasing T1 by 100oC or decreasing T2 by 100oC, would result in a greater improvement in the efficiency of the engine?
Efficiency of Carnot engine,
(i) When T1 is increased form 1500oC to 1600oC or 1600 + 273 = 1873 K and T2 remains constant i.e., 500oC or 500 + 273 = 773 K, we have
(ii) When T1 remains constant i.e., 1500oC or 1500 + 273 = 1773 K and T2 is decreased by 100oC from 500oC to 400o or 400 + 273 = 673 K, we have
Thus, η2 > η1. Hence efficiency will increase if T2 is decreased from 500oC to 400oC.
|