Question 1: A helium balloon consist of a large bag loosely filled with 600 kg of helium at an initial temperature of 10 degrees Celsius. while exposed to the heat of the sun, the helium gradually warms to a temperature of 30 degrees Celsius. the heating proceeds at a constant pressure of 1.0atm. how much heat does the helium absorb during this temperature change?Solution heat absorbed at constant pressure = >mCp*dTmass of helium = 600 kgCp helium = 5.19 kJ/kgKdeltaT = >30 -10 = 20Q => heat absorbedQ => 600 x 5.19 x 1000 x 20Q = 62280 kJ
Question 2: A large closed bag of plastic contains 0.10 m^3 of an unknown gas at an initial temperature of 10 degrees Celsius and at the same pressure as the surrounding atmosphere, 1.0 atm. You place this bag in the sun and let the gas warm up to 38 degrees Celsius and expand to 0.11m^3. During this process, the gas absorbs 840 cal of heat. Assume the bag is large enough so the gas never strains against it, and therefore remains at a constant pressure of 1.0atm.A) how many moles of gas are in the bag?PV=nRT
1*0.1 = n *8.314*283n = 0.1/8.314*283= 4.25*10^-5 mol.
B) what is the work done by the gas in the bag against the atmosphere during expansionw=PextΔV = 1* ( 0.11 – 0.10)= 0.01 JC) What is the change in the internal energy of the gas in the bag*same as work done*
D) Is the gas a monoatomic gas? A diatomic gas?Diatomic, pressure remains constant=> characteristic of a diatomic molecule/gasQuestion 3: A Carnot engine has 40g of helium gas as its working substance. it operates between two reservoirs of temperatures T1 = 400 degrees C and t2 = 60 degrees C. The initial volume of the gas is v1 = 2.0 x 10^-2 m3, the volume after isothermal expansion is v2 = 4.0 x 10^-2 m3, and the volume after the subsequent adiabatic expansion is v3 = 5.0 x 10^-2 m3.a. Calculate the work generated by this engine during one cycle.b. Calculate the heat absorbed and the waste heat ejected during one cycle.Solution(a)…….. mass of the He gas, m => 40 g
T1 = 400°C => 673 K
T2 = 60° C => 333 K
v1[initial vol.] = 2 x 10-2 m³
vol. of the gas after isothermal expansion, v2 = 4 x 10-2 m³
vol. of the gas after adiabatic expansion, v3 = 5 x 10-2 m³
no. of mol, n = m /mm => 40 /4 = 10
but, for an isothermal expansion;
work done by a carnot engine is given as,
Wiso = n R T log (v2 / v1)……………..(eq. 1)
where, R = gas constant => 8.314 J/mol.K
T = initial temperature = 400° C
inserting all these values in above eq.
Wiso = (10) (8.314 J/mol.K)*(400 0C)*log [(4 x 10-2m3)/(2 x 10-2 m3)]
Wiso => 23046.4 J
For an adiabatic expansion :
work done by a carnot engine is given as,
Wad = n R (T2 – T1) / (k – 1)…………….. (eq. 2)
where, k = 1.4 <=> for monoatomic gas
Wad = (10)*(8.314 J/mol.K)* [(333 K) – (673 K)]/(1.4 – 1)
Wad = – 3040 J
(b)………
DeltaW = Wiso + Wad………….. ( eq. 3)
inserting the value in eq.3
DeltaW => (23046.4 J) – (3040 J)
DeltaW = 20007 J
DeltaW = 2 x 104 J
Question 4: On a hot day a house is kept cool by an air conditioner. The outside temperature is 32°C and the inside temperature is 21°C. Heat leaks into the house at the rate of 9000 kcal/h. If the air conditioner has the efficiency of a Carnot engine, what is the mechanical power that it requires to hold the inside temperature constant?Solution Carnot Efficiency = 1 – Tlow / ThighCarnot eff => 1 – (273+21)/(273 + 32) = 0.036For maintaining temperature, Heat rejected => 9000 kcal / hrEfficiency => Work / (Work + Heat rejected)0.036 = Work / (Work + 9000)Work = 336.7 kcal/hr = 336.7 kcal / (60 × 60) s = 0.0935 kcal/s = 391 W
Question 5: In a nuclear power plant, the reactor produces team at 520 degrees C and the cooling tower eliminates waste heat into the atmosphere at 30-degree C. The Power generates 500 megawatts of electric or (mechanical) power.
A. If the efficiency is that of a Carnot engine, what is the rate of release of waste heat (in megawatts). Solution For carnot engine efficiency => n = 1 – T2 /T1 => Power generated (W) /heat supplied (Q1)T2 = lower temp = 30°C => 273K + 30 = 303KPower gen. => W = 500*106 WQ2 = heat vented out = Q1 – Wfrom the above eqn. = n = 1 – (303/793) = (1500*106)/Q1==> 0.6179Q1 = 1500*106Q1 = 2427.55MW
B. Actual efficiencies of nuclear power plants are about 33%. For this efficiency, what is the rate of release of heat?
Rate of release;33/100*2427.55 = 801.092 MW per hrRate in sec. ;801.092/3600 = 0.2225MW per sec
