Question 444.
Figure (a) shows 2 metal bars X and Y which have the same cross-sectional and length joined in series. The thermal conduction of Y is three times of X.
a. If the rods are well lagged, find the temperature at the common junction.
b. The rods are then joined in parallel as shown in figure(b). Compare the rate of heat flow in figure(b) with that in figure(a) if the rods are well-lagged in both cases.
Answer 444.
a) Let T = temperature at the common junction.
Heat flow rate through both the blocks will be the same
=> Q/t = kA(T - 0)/L = 3kA(100 - T)/L
=> T = 300 - 3T
=> T = 75° C.
b)
Part a) is a case of series connection and that of b) is of parallel connection.
For part a),
Equivalent thermal resistance = L/kA + L/3kA = 4L/3kA
For part b),
equivalent thermal resistance = 1 / [1/(L/kA) + 1/(L/3kA)] = 1 / (4kA/L) = L/(4kA)
=> For a),
(Q/t)_a = 100 / (4L/3kA) = 75kA/L
and for b)
(Q/t)_b = 100 / (L/4kA) = 400kA/L
=> ratio of heat flow rate of b) to a)
[400kA/L) / (75kA/L)
= 16/3.
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Question 420.
Calculate the work, heat, delta U, delta H for the reversible isothermal expansion of 2 moles of ideal gas from 16 L to 95 L at 0 degrees celcius.
Answer 420.
For isothermal expansion, temperature being constant,
Change in internal energy, ΔU = 0.
Work done by the ideal gas in reversible isothermal process, W
= nRT ln(V2/v1)
= 2 * 8.31 * 273 * ln(95/16) J
= 8082 J.
Heat absorbed by the gas, Q
= work done by the gas
= 8082 J
= 8082 * 0.239 calories
= 1932 calories.
ΔH = ∫(T1 to T2) Cp dT = 0 as T1 = T2.
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Question 397.
A gas is suddenly compressed to 10 times its original pressure . Calculate the rise in temperature of gas if its initial temperature is 27 degree C . take γ = 1.5.
Answer 397.
Sudden compression means there is hardly any time for heat exchange to occur with the environment which indicates that the process is adiabatic for which
PV^γ = constant
Also, for an ideal gas, PV = nRT => P^γ * V^γ = (nR)^γ * T^γ = constant * T^γ
taking ratio of these two eqns.,
P^(γ - 1) = (constant) * T^γ
=> (T2/T1)^γ = (P2/P1)^(γ - 1)
=> (T2/300)^(1.5) = (10)^(1.5 - 1)
=> T2 = (300) * (10)^(1/3)
=> T2 = 646 K
=> T2 = 373° C
=> rise in temperature of gas
= 373 - 27
= 346° C.
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Question 375.
Initially 0.800 mol of an ideal gas in a container occupies a volume of 3.10 l at a pressure of 3.90 atm with an internal energy U1 = 364.8 J. The gas is cooled at a constant volume until its pressure is 2.50 atm. Then it is allowed to expand at constant pressure until its volume is 6.20 l. The final internal energy is U2 = 467.7 J. All processes are quasi static. Draw this process on a PV diagram. What is the work done by the gas? What is the heat absorbed by the gas?
Answer 375.
The processes are shown on PV diagram as under.
Work done by the gas in isochoric process = 0
Work done bt the gas in isobaric process = PdV
= (2.50) * (6.20 - 3.10) lit-atm
= 7.75 lit-atm
= 785.26875 joule ... [Refer to the link - http://www.convertunits.com/from/L+*+atm…
Heat absorbed by the gas = Increase in internal energy + work done by the gas
= 467.7 - 364.8 + 785.3 joule
= 888.2 joule
= (888.2) * (0.239) calories
= 212.3 calories.
[No. of moles of the gas is not needed in any of the above calculations.]
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Question 370.
A cylinder/piston setup contains air at 100kPa and 20° C and has a volume of 0.3 m^3. The air is compressed to 800 kPa in a reversible process in which PV^1.2 is held constant, after which it is expanded back to 100 kPa in a reversible adiabatic process.
Calculate the final temperature and the net work.
Answer 170.
Given equation PV^1.2
=> P1V1^1.2 = P2V2^1.2 ... ( 1 )
According to Ideal Gas Law equation,
PV = nRT
=> P1V1/T1 = P2V2/T2
=> P1V1T2 = P2V2T1
=> (P1V1T2)^1.2 = (P2V2T1)^1.2 ... ( 2 )
From ( 1 ) and ( 2 ),
P1^0.2 * T2^1.2 = P2^0.2 * T1^1.2
=> T2 = T1 * (P2/P1)^0.2/(1.2)
=> Final temperature,
T2 = (20 + 273) * (8)^(1/6) = 414.4 K = 141.4° C
Work done, W
= RT1 / (n - 1) * [1 - (p2/p1)^(n-1)/n]
= (8.314) * (293) * [1 - (8)^1/6)
= - 1009 J
Work done in adiabatic expansion, W'
= RT2 / (γ - 1) * [1 - (p2/p1)^(γ-1)/γ] ... γ = 1.41 for air
= (8.314) * (414.4) * [1 - (1/8)^(0.41/1.41)]
= 1563 J
Net work done
= - 1009 + 1563 J
= 554 J.
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Question 356.
A rigid tank contains 20 lbm of air at 50 psia and 80° F. The air is now heated until its pressure doubles. Determine: (a) the volume of the tank and (b) the amount of heat transfer.
Answer 356.
R = 1545.35 ft-lbf / (lbmol-Rankine)
(http://www.katmarsoftware.com/gconvals.h…
PV = nRT
P = 50psia = 50 * 144 lbf/sqft abs.
=> 50* 144 V = (20/28.9644) * 1545.35 * (80+460)
=> V = 80.0 cu. ft.
20 lb mass
= 20 * 453.4 g
Cv = 5 cal / g-mol - K
When preesure doubles with volume constant, temperature doubles
Temp. = 80 degrees F = (80-32) / 1.8 degrees C = 26.7 degrees C = 299.8 K
=> increase in temperature = 299.8 K
Heat absorbed = change in internal energy as work done is zero at constant volume
=> Heat absorbed
nCv dT
= (20 * 453.4/28.966) * 5 * 299.8 cal.
= 469272 cal.
= 469272/252 Btu
= 1862 Btu.
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Question 324.
1) A reversible Carnot cycle heat engine operates between an upper temperature T₁ and a lower temperature T₂. It consists of four steps that take the engine through the cycle: (T₁, V₁)→ (T₁,V₁ʹ)→(T₂, V₂)→(T₂, V₂ʹ)→ (T₁, V₁). The volumes in the cycle satisfy the relation: ( V₂ʹ - b)/ (V₂- b)= (V₁- b)/(V₁ʹ - b). Compute the work w for the entire cycle.
2) The working substance of the engine in part 1 is one mole of a gas with equation of state, P(V-b)= RT, where b is a constant. Compute the heat absorbed by the system in the process (T₁, V₁)→ (T₁,V₁ʹ).
Answer 324.
Carnot cycle is made up of four reversible processes of an ideal gas
i) isothermal expansion,
ii) adiabatic expansion,
iii) isothermal compression and
iv) adiabatic compression.
1)
Net work done in adiabatic processes is zero.
Net work done in isothermal processes
= W₁ + W₂
= RT₁ ln[(V₁' - b)/(V₁ - b)] + RT₂ ln[(V₂' - b)/(V₂ - b)]
= RT₁ ln[(V₁' - b)/(V₁ - b)] - RT₂ ln[(V₁' - b)/(V₁ - b)]
= R(T₁ - T₂) ln[V₁' - b)/(V₁ - b)]
which is the net work done for the entire cycle.
2)
Heat absorbed in the process (T ₁, V₁)→ (T ₁,V₁ʹ).
Q
= nCvdT + W₁
= 0 + RT₁ln[V₁' - b)/(V₁ - b)].
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Question 255.
Four liters of a monatomic ideal gas, originally at 470 K and a pressure of 3 atm (called state 1), undergoes the following processes:
1 → 2 isothermal expansion to V2 = 7V1
2 → 3 isobaric compression
3 → 1 adiabatic compression to its original state
Find the pressure, volume, and temperature of the gas in state 2 and 3.
Answer 255.
4 liters at 470 K and 3 atm
= 4 * (273/470) * (3/1) liter at NTP
= 6.970 liter
22.4 liter at NTP = 1 mole of gas
=> quantity of gas
= (6.970/22.4) = 0.311 mole.
Gas in State 2:
V2 = 28 liter
P2 = 3/7 atm
T2 = 470 K
Gas in State 3:
P3 = 3/7 atm
Considering adiabatic compresion 3 to 1,
P1V1^γ = P3V3^γ
=> V3
= V1 * (P1/P3)^(1/γ)
= 4 * (3/(3/7)^(1/1.4)
= 16.06 liter
Also,
P1V1/T1 = P3V3/T3
=> T3
= T1 * ((P3V3) / (P1V1)
= 470 * (3/7) * (16.06) / [3 * 4)
= 269.6 K.
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Question 230.
A reversible heat engine receives heat from two thermal reserviors at 870K and 580K,and rejects 50KW of heat to a sink at 290K.If the engine output is 85KW,make calculations for the energy efficiency and heat supplied by each reservior.
Answer 230.
Let Q1 = heat supplied by the first reservoir
and Q2 = heat received by the 2nd reservoir
=> Q1 + Q2 = work done + heat rejected
=> Q1 + Q2 = 85 + 50 = 135 kW ... (1)
Efficiency of the engine for heat received from the first source
= (870 - 290) / (870) = 2/3
Efficiency of the engine for heat received from the 2nd source
= (580 - 290) / (580) = 1/2
=> (2/3)Q1 + (1/2)Q2 = 85
i.e., 4Q1 + 3Q2 = 510 ... (2)
Solving eqns. (1) and (2),
Q1 = 105 kW and
Q2 = 30 kW
Efficiency
= W / (Q1 + Q2) x 100
= 85 / (105 + 30) x 100
= 62.96 %.
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Question 223.
Suppose that in an insulated container, 0.100 kg of water at 20.0 degrees celcius is mixed with 1.500 kg of ice at -15.0 degrees celcius. Find the final temperature of the system.
Answer 223.
1)
100 g of water will give up 2000 cal of heat in cooling from 20° C to 0 °C.
2)
1500 g of ice will need 1500 * 0.5 * 15 = 11250 cal of heat to get heated up from
-15 °C to 0 °C
3)
Since heat needed by subcooled ice is much more than what water can give by cooling down to 0 °C, some water will have to freeze to ice to provide that heat.
4)
2000 cal of heat is given by water cooling to 0 °C and further 9250 cal of heat is needed for entire quantity of ice to get heated to 0 °C. 100 g os water can give 80 * 100 = 8000 cal of heat if it waere to freeze to ice at 0 °C.
5)
From the above observations, it is clear that the quantity of ice which is subcooled to - 15 °C will not heat up to 0 °C as the heat available from 100 g water will be insufficient. Suppose, ice gets heated up to - T °C, then
heat needed by 1500 g of ice = 1500 * 0.5 * (15 - T)= 750 (15 - T)
6)
With heat balance in the above situation,
750 (15 - T) = 2000 + 8000 + 100 * T
=> 11250 - 750T = 10000 + 100 * 0.5 *T
=> 800T = 1250
=> T = 1.5625 °C
That means the final temperature of the mixutre will be - 1.5625 °C and the entire mixture will be subcooled ice at - 1.5625 °C ======================================
Verification:
Heat lost by 1500 g of ice = 1500 * 0.5 * (15 - 1.5625) = 10078 cal
Heat gained by water at 20 °C to get frozen to subcooled ice at - 1.5625 °C
= 100 * (20 + 80 + 0.5 * 1.5625) cal = 10078 cal.
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Question 97.
Mr Smith has a 3kW. water heater containing 120kg. of water initially as 20°C, and a clock radio which consumes 5W. No other appliances are using any electricity. There is 41p. of credit remaining on his electricity meter.
The alarm is set to sound after 8 hours, but if the meter runs out sooner than that then Mr Smith will oversleep.
What is the hottest temperature that his thermostat could be set to without running the electricity meter out of credit?
Specific Heat Capacity of water = 4200 J/(kg.K), Price of electricity = 8p/kWh.
Assume the water heater is thermally insulated, and neglect its own heat capacity.
Answer 97.
Let T = the maximum temperature to which the thermostat is set, then the temperature of water should not reach this value in 8 hours.
In 41 p, amount of electricity available
= 41/8
= 5.125 kWh
Of this, amount of electricity consumed by clock
= 5 x 8
= 40 Wh
= 0.040 kWh
Balance available for consumption by water heater
= 5.125 - 0.040 kWh
= 5.085 kWh
= (5.085) * (3600000) J
This should equal the energy for heating 120 kg of water
=> 120*(T - 20)*(4200) = (5.085) * ( 3600000)
=> T - 20 = (5.085) * (3600) / [(12) * (42)]
=> T - 20 = 36.32
=> T = 56.32° C.
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Question 42.
An insulated beaker with negligible mass contains a mass of 0.345 kg of water at a temperature of 60.4 deg. Celsius.
How much ice at a temperature of -11.2deg. Celsius must be dropped in the water so that the final temperature of the system will be 20.2deg. Celsius?
Take the specific heat for water to be 4190 J/(kg*K), the specific heat for ice to be 2100 J/(kg* K), and the heat of fusion for water to be 334 kJ/kg.
Answer 42.
Let the mass of ice required = x kg.
Now, heat lost by 0.345 kg water when its temperature falls from 60.4 to 20.2 degrees
= 0.345 *1000*( 60.4 - 20.2 ) = 13869 cal.
Heat gained by x kg of ice when its temperature rises from - 11.2 to 0 degree
= x*1000*0.5*11.2 = 5600x cal.
Heat gained by x kg of ice at 0 degree to melt to water at 0 degree
= x*1000*80 = 80000x cal
Heat gained by x kg of water when its temperature rises from 0 to 20.2 degrees
= x*1000*1*20.2 = 20200x cal.
Total heat gained by ice
= ( 5600 + 80000 + 20200 ) x cal.
= 105800 x cal.
Heat gained by ice = heat lost by water
=> 105800x = 13869
=> x = 0.131 kg.
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