| Problem No 11.1 |
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Prove that the Celsius anf
Fahrenheit scales have the same reading at -40º ? |
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| Datas |
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ºC = |
-40 |
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ºF = |
-40 |
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To prove ºC = ºF |
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We know from the relation of Celsius
and Fahrenheit scale that, |
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ºC = [(ºC x 9/5 ) + 32 ] ºF |
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putting values we get |
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-40ºC = [(-40º x9/5) +32] ºF |
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-40ºC = [(-72+32]ºF |
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-40ºC
= -40 ºF |
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This -40º is the point where both
scales have the same readings. |
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| Problem
No 11.2 |
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A
test tube filled with air is sealed off S.T.P. It is then heated to 300
ºC.What then is the air |
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pressure
in the tube in Newton per square meter, in pascles, and in atmosphere? |
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Given 1 atmosphere = 76 cm Hg = |
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= 1.01 x Pa. |
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| Datas |
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Initial Temperature = T1 = |
0ºC |
273 K |
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Initial Tpressure =
P1 = |
1 atm |
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Final Temperature =
T2 = |
300ºC |
573 K |
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Initial Volume =
V |
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Final Pressure =
P2 = ? |
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From the general gas equation |
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P1V1 / T1
= P2V2 / T2 |
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Since V1 = V2 = V |
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P1 / T1 = P2 / T2 |
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P1T2 / T1
= P2 |
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P2 = |
1 x 573 |
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273 |
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P2 = |
2.098901 |
atm Ans |
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P2 = |
2.12 x |
Pascals |
Ans |
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P2 = |
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N/m² |
Ans |
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| Problem
No 11.3 |
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Air at 27ºC and atmosphere pressure
is suddenly compressed to one twentieth of its original |
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| volume and pressure of 30 atmosphere. What
is the temperature of the gas? |
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Datas |
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Initial temperature of the gas= |
T1 = |
27 |
ºC = |
300 K |
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Final temperature of the gas= |
T2 = |
? |
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Initial Pressure = |
P1 = |
1
atm |
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Final Pressure = |
P2 = |
30 atms |
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Initial Volume = |
V1 = |
V |
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Final Volume = |
V2 = |
V/20 |
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P1 V1 / T1
= P2 V2 / T2 |
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T2 = |
P2 V2
T1 |
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P1 V1 |
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T2 = |
30 x V/20 x 300 |
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1 x V |
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T2 = |
30 x300 xV |
T2 = |
9000 |
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1 x 20 x V |
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20 |
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T2 = |
450 - 273 |
= 177 |
°C Ans |
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| Problem
No 11.4 |
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An
electric heater supplies 1800 W of power in the form of heat to a tank of
water. How |
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| long
will it take to heat the 200 kg of water in the tank from 10ºC to 70ºC?
Assume heat losses to |
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| surrounding
to be negligible. Specific heat of water is 4200 j/kg K. |
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Datas |
Power of heater = |
P =1800 W = 1800 J/s |
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Mass of water = |
m =200kg |
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Initial temperature = |
T1=10°C |
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Final temperature = |
T2=70°C |
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Specific heat of water = |
C=4200 j/kg K |
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Time required = |
h = ? |
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Energy = Power x Time |
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Q = Pt |
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Q = 200kg X 4200j/kg °C x (70°C -
10°C) |
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Q = 840000 J x 60 |
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Q =50400000 J |
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t = Q / P |
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t = |
50400000 |
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1800 |
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t = |
28000 |
sec |
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t = |
7.777778 |
hrs |
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| Problem
No 11.5 |
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A copper calorimeter has a water
equivalent of 5.9g. That is in heat exchanges, the |
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| calorimeter behaves like 5.9g of water, It
contain 40g of oil at 5.0ºC, when 100g of lead at 30ºC |
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| is added to this, The final temperature is
48.0C. What is the specific heat capacity of oil? |
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Specific heat of lead =128.1 j/kg ºC. |
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Datas |
Mass of water in calorimeter = m1
= |
5.9g = |
0.0059kg |
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Mass of oil = m2 = |
40g = |
0.04kg |
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Mass of lead =
m3 = |
100g= |
0.1 kg |
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Initial temperature of
Calorimeter = T1 = |
50°C= |
323 K |
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Initial temperature of oil = T2 = |
50°C= |
323 K |
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Initial temperature of oil = T3 = |
30°C= |
313 K |
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Specific heat of calorimeter =
C1 = |
4200 J/kg K |
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Specific heat of oil = C2
= |
? |
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Specific heat of Lead = C3
= |
128 J/kg K |
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Final temperature of the mixture
= T
= |
48°C = |
48+273 = |
321 K |
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According to the law of
heat exchange |
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Heat lost by the hot body
= Heat gained by the cold body |
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m1C1(T1 - T) + m2C2(T2 -T) = m3C3(T -T3) |
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m2C2(T2 -T)
= m3C3(T -T3) -m1C1(T1 - T) |
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C2 = m3C3 (T - T3) - m1 C1 (T1 -T) |
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m2 (T2 -T) |
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C2 = 0.1 x 128 x (321-313) - 0.0059 x 4200
x (323 -321) |
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0.04 x (323 -321) |
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C2 = 230.4 - 49.56 |
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0.08 |
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C2 =2260.5 J/kg K |
Ans |
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| Problem
No 11.6 |
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One
meter steel bar is correct at 0ºC and another at 2.5ºC. What is the
difference between |
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| their
lengths at 20ºC? |
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Datas |
Initial length of first steel
bar = |
L1 = |
1 m |
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Initial Temperature of first steel
bar = |
T1 = |
0°C |
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Final Temperature of first steel
bar = |
T2 = |
20°C |
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Initial length of second steel
bar = |
L2 = |
1 m |
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Initial Temperature of second steel
bar = |
T3 = |
2.5°C |
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Final Temperature of second steel bar = |
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T2 = |
20°C |
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Coefficient |
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Final length of first steel bar = |
L3 = |
? |
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Final length of second steel
bar = |
L4 = |
? |
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Length difference of both Bars =L4 - L3 = |
? |
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L3 = L1( 1+ T )
= L1[ (1+ (T2 - T1)] |
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L3 = 1[1+
x (20°K - 0°K) |
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L3
=1.00024 m |
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| L4 = L2( 1+ T ) = L2[ (1+ (T2 - T3)] |
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L4 = 1[1+
x (20°K - 2.5°K) |
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L4 = 1[1+0.00021 |
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L4 = 1.00021m |
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Length difference of both Bars =L4 - L3 = |
? |
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Length difference of both Bars =L4 - L3 = |
1.00024m - 1.00021m |
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Length difference of both Bars =L4 - L3 = |
0.00003m = .003cm Ans |
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| Problem
No 11.7 |
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What is the change in internal
energy of 200g of Nitrogen as it is heated from 10°Cto 30°C |
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| at constant
Volume? |
For Nitrogen Cv = 0.177 Cal/ g°C |
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Datas |
Mass of Nitrogen = m = 200g
= 0.2 kg |
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Initial temperature of N2
= T1 = 10°C
= 283 K |
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Final temperature of N2 = T2 = 30°C = 303 K |
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Cv for N2 |
= 0.177
cal/g °C |
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Change in the internal energy =
U |
= ? |
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U = m Cv T |
=
200 x 0.177 x (30 -10) cal |
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= 35.4 x 20
Cal |
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= 708 cal |
=
708 Cal x 4.185 j / Cal |
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= 2960 J |
Ans |
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| Problem
No 11.8 |
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| Five grams of helium gas is heated from
-30 ˚C to 20 ˚C. Find the change in its internal energy |
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| and the
work done by the gas if heating occurs at (a) constant volume, (b) constant
pressure. For |
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| He Cv =0.75
cal/(g)(˚C) and Cp = 1.25 cal/(g)(˚C). |
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| datas |
mass of He m = |
5 |
gm = |
0.005 |
kg |
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Molecular mass M = |
4 |
kg/kmol |
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No of kilomoles n = |
.005kg = |
0.00125 |
kmol |
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4kg/kmol.K |
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T1 = |
-30 |
˚C = |
243 |
K |
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T2 = |
20 |
˚C = |
293 |
K |
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∆T = |
50 |
K |
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Cv = |
0.75 |
Cal/kmol K = |
12540 |
j/kmol K |
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Cp = |
1.25 |
Cal/kmol K = |
20900 |
j/kmol K |
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∆U = |
? At
constant Volume & constant pressure |
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∆W = |
? At
constant Volume & constant pressure |
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| (a) |
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∆U = |
? At
constant Volume i.e Cv |
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∆U = |
nCv∆T |
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∆U = |
.00125kmol x 12540j/kmol K x 50 K |
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∆U = |
783.75 |
joul. Ans |
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∆W = |
? At
constant Volume i.e Cv |
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∆W = |
P∆V |
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∆W = |
P x 0 |
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∆W = |
0 Ans |
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| (b) |
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∆U = |
? At
constant pressure i.e Cp |
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∆U = |
will be same because ∆T is
same i.e there is no temperature difference |
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∆U = |
783.75 |
joul. Ans |
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∆W = |
? At
constant pressure i.e Cp |
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∆W = |
P∆V |
= nR∆T |
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As PV =nRT |
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∆W = |
nR∆T |
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∆W = |
.00125kmol x (Cp-Cv) x 50 K |
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Cp - Cv = |
8360 |
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∆W = |
.00125kmol x 8360j/kmol K x 50 K |
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∆W = |
522.5 |
joul.
Ans |
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| Problem
No 11.9 |
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| Nitrogen gas is adiabatically compressed
in such a way that its temperature is to rise from 20 ˚C to |
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| 500
˚C. To what fraction of its original volume must the gas be
compressed? |
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Initial temperature of N2 T1 = |
20 |
˚C = |
293 |
K |
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final temperature of N2 T2 = |
500 |
˚C = |
773 |
K |
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Initial Volume V1
= |
V |
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V2 / V1 = |
? |
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Heat supplied ∆Q = |
0 |
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Because gas is adiabatically
compressed |
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|
r-1
. |
r-1 |
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T1 V1 = |
T2 V2 |
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we know r = Cp/Cv |
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r = gamma |
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for Nitrogen r = |
1.4 |
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After
eliminating the T from eq 11.28 by using the ideal gas equation…….and |
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| using
the simplified equation from example No 11.10. we have, r-1 |
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1.4-1 |
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(V2
/V1 ) = |
T1 / T2 |
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(V2 /V1 )
= |
T1 / T2 |
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0.4 . |
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(V2 /V1 )
= |
293 K / 773 K |
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1/.4 |
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(V2 /V1 )
= |
(294 K / 773 K) |
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|
. 2.5 |
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(V2 /V1 )
= |
0.379043 |
) |
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(V2 /V1 )
= |
.38 x .38 x √.38 |
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As |
√.38 = |
0.6164 |
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(V2 /V1 )
= |
0.143673 |
x √.38 |
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(V2 /V1 )
= |
0.143673 |
x .6164 |
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(V2 /V1 )
= |
0.08856 |
Ans |
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| Problem
No 11.10 |
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| In a high pressure steam turbine engine, the
steam is heated to 600 ˚C and exhausted at about 90 ˚C. |
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| What is
the highest possible efficiency of any engine that operates between these two
temperatures? |
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|
Initial temperature T1 = |
600 |
˚C = |
873 |
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final temperature T2 = |
90 |
˚C = |
363 |
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∆T = |
510 |
K |
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Efficiency of
engine η = |
1-T2 /T1 |
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Efficiency of
engine η = |
1- 363K/873K |
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Efficiency of
engine η = |
0.584192 |
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Efficiency of
engine η = |
58.41924 |
% Ans |
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| Problem
No 11.11 |
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| Temperature diference between the
surface water and the bottom water in manicher lake might be |
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| 5
˚C.Assuming the surface water to be at 20 ˚C, what is the highest
possible efficiency a steam engine |
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| could have
if it operates between these two temperatures? |
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|
Temperature of top T2 = |
15 |
˚C = |
288 |
K |
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Temperature of bottmT1 = |
20 |
˚C = |
293 |
K |
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Difference of temp ∆ T = |
5 |
˚C = |
5 |
K |
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Efficiency of Carnot Engine
= |
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η = |
1- T2 / T1 |
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η = |
1- 293 K / 278 K = |
1 - |
0.982935 |
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η = |
0.017065 |
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η = |
1.706485 |
% Ans. |
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| Problem
No 11.12 |
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| A Carnot engine whose low temperature
reservoir is at 7˚C has an efficiency of 40%. It is desired to |
| increase
the efficiency to 50%. By how many degrees must that temperature of the high
temperature |
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| reservoir increase? |
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Initial temperature T1 = |
? |
˚C = |
? |
K |
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final temperature T2 = |
7 |
˚C = |
280 |
K |
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η1 = |
40% |
=.4 |
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η2 = |
50% |
=.5 |
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Efficiency of
engine η = |
1-T2 /T1 |
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T2 /T1
= |
1-η |
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T2 = |
(1-η)T1 |
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(i) When Efficiency is 40% |
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T2/T1
= |
(1-η1) |
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280K/T1
= |
1-.4 |
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280K/T1
= |
0.6 |
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T1 /280 = |
1/.6 |
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T1 = |
1/.6 x 280K |
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T1 = |
1.667 |
x 280K |
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T1 = |
466.7 |
K |
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(ii) When Efficiency is 50% |
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T2/T1
= |
(1-η2) |
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280K/T1
= |
1-.5 |
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280K/T1
= |
0.5 |
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T1 /280 = |
1/.5 |
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T1 = |
1/.5 x 280K |
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T1 = |
2 |
x 280K |
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T1 = |
560 |
K |
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Required increase in temperature is= |
560K - 467K = |
93.33 |
K |
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|
| Problem
No 11.13 |
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What is the change in the entropy of
30g of water at 0˚C as it is changed into ice at 0˚C. |
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Heat of fusion for ice is; Hf = 336000 J/kg |
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Temperature of ice = T = |
0 |
˚C = |
273 |
kg |
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mass of water = m = |
30 |
gm = |
0.03 |
kg |
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Change in entropy ∆S = |
? |
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Heat of fusion for ice = Hf = |
3E+05 |
|
J/kg |
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∆Q = |
mHf |
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∆Q = |
10080 |
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J |
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Change in entropy ∆S = |
∆Q/T |
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Change in entropy ∆S = |
10080J/273K |
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|
Change in entropy ∆S = |
36.92 |
J/K |
Ans |
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| Problem
No 11.9_Old Book |
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|
|
An airconditioner absorbs heat from
its cooling coils at 10°C and expels heat to the outside at |
| 40°C.
Calculate the maximum coefficient of performance of the air conditioner. |
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| Datas |
Low temperature = |
T2 = |
10°C = |
273 +10 = |
283 K |
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High temperature = |
T1 = |
40°C = |
273 +40 = |
313 K |
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E = |
_T2_ |
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T1 - T2 |
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E = |
283 |
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313 - 283 |
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E = |
283/30 |
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E = |
9.433333 |
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|
