Question

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A Carnot refrigerator removes 10 kW of heat from a thermal reservoir at 3°C, using 4 kW of power. What is the temperature of the high-temperature thermal reservoir? O a. 4.2°C O b.2.14°C O c. 113°C d. 145°C How much power is needed to operate a Carnot heat pump if the pump receives heat at 10°C and delivers 50 kW of heat at 40°C? O a. 37.5 kW O b. 5.30 kW O c. 150 kw O d. 4.79 kW What is a frictionless machine that can operate indefinitely? O a. Perpetual Motion Machine - type 2. O b. Possible. O c. Perpetual Motion Machine - type 1. 0.50 kg of air is heated at constant pressure from 25°C to 100°C. The source of the heat is at 200°C. What is the entropy generation for the process? O a. 0.301 kJ/kg-K O b. 0.617 kJ/kg-K O C. 1.24 kJ/kg-K O d. 0.0331 kJ/kg-K A solid block of metal with a specific heat of 0.52 kJ/kg-K and a mass of 3 kg is cooled from 80°C to 20°C. What is the change in entropy of the block? O a. -0.291 kJ/kg-K O b. -0.872 kJ/K O C.-0.721 kJ/kg-K O d. -2.16 kJ/ K If 20 kW of heat is added to a system at 50°C. What is the entropy transport for the process? O a. 0.40 kW/K O b. -0.40 kW/K O c. 2.5 kW/K O d. 0.0619 kW/K

Answer 1

Th= Lower temp. = 3°= 276K THE Highen temp. 4kw Q=lokoh G 176KT lower temp. = heat taken from ar Reservoir = lokw wa power input = 4 KW Now, we know for Cannot Refrigeratori Cop) = Cofficient of performance of Ref. Refrigerator. = Desired effect = 10 kw = 2.5 Work in Dut y kw - Also, (Cop) oef. = I 2.5= 276 26 - 276 - TTH=386.4K . = 113.4°C Ary [TH=313K y = 50kw t Son 2) T = 10'c = 283k THE 40 c=313K W= power input Now, For Cannot heat pumpir = (Cop) = Desiered effect 50 power input QL (1=2836]

6 Also (COP) ip = I 5 ? 5 313 → W = 4.992 kw thy 5 Son 3). Perpetual motion machine of 1st kind.- It is a hypothetical machine that continues ndefinately without any external source of energy. 9t is impossible in practice because Of friction. perpetual motion of and Kindi It is hypotretical machine that derives its energy from a source at a lower temp. It is impossible in practice because of - 2nd law of thermodynamics. So, Ang af given question is :- Perpetual motion machine type 1 Ang - Note: perpectual motion machine is not possible in real life. Source T= 2ooc =4.73K m=0.5kg, heat at constant " pressure from, T,= 25c = 298K to Tzê 100°c =373K (cp) = 1005 kJ Kg-k

a= Amt, of heat needed to heat air from 25°c to loo'c = m Cp (T2-T) = 0.5x2005 (100-25) = 37.6875 KJ | Now, Sgon. = entropy generation & kes air + les source = mcp in (17 2 ) tece = 0.5X105 in (33) + (-37-45.25) 30.033) KIK Ans. source son s. m = 3kg (p=0.52 k ok I TE 80'c = 353k, T2 = 20°c = 293K E (AS) block = m Cp in Iz = 380:52 in (2013) =-0.291 kg Ang T=50c=323k Soln 6). @ = 2oko (added) entropy increase of System Entropy transport =

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