Heat numericals for XII year

Q.1 Q.2 (c) 1200 J (🔺️Q) of heat energy is supplied to the system at constant pressure. The intermal energy of the system is increased by 750J (🔺️U) and the volume by 4.5 cubic meters (🔺️V), Find the work done (🔺️W) against 

1200] (AQ) of heat  energy is supplied to the system at constant pressure. The internal energyof the  system is increased by 750] (AU) and the volume by 4.5 cubic meters (AV). Find the work (4) d on e (AW =?) against piston and the pressure (P =?)onthe piston. P=? AW =?, Hint- AQ = 1200/,AU =750,AV = 45m®, (i) (ii) AQ = AU + AW @ =  1200 = 750 + AW =  AW = 1200 — 750 = [AW =450] AW = PAV = 45 0=Px 45 450 p Sp =—— 4.5 = |P = 100 N/m? 1994 (4507, 100 N/m?) A meter  bar  (Lo, = 1m) of steel is  at 0°C (T, = 273 K) and another meterbar(Lo,=1m)isat -2.50C (T/ = 270.5 K). Wh at will be the difference between their lengths (AL = L, — L, =?) at (6) 30°C (T, = 303K). (@=12x1075 /K) T, =0°C = 0+ 273 = 273K, Lo, =1m, Hint- Lo, =1m, T/ 25°C = -25+273 =2705K, 30 3= AL=1 L, — L, =? T, =  30°C =3 0+ 273 = 303K, Ly = Lo, (1 + aAT) SL =Lo[1+ a(T,— T)] &g t ; L, =1[1+12x107° (303 — 273)] L,=1[1+12x10"x30] SL =1[1+ 36x107% =L; =1x1.0036 =  |L; = 1.00036 m L, = Lo, (1 + aAT) = Ly=lo,[1+ a (T, 1) &g t ; L, =1[1+12x107° (303 — 270.5)] = &g t ; L, =1[1+1 2 x 107° x325] =L,=1[1+39x107] =L, =1x 1.0039 =  |L, = 1.00039 m AL=1 L, — L, =  AL = 1.0039 — 1.0036 =[AL =0.00003m=30x10"m 1995 ( 0 .00003 m OR 30 x 10°m) A heat engine performs work at the rate of 50 kW (P = 50,000 W). The efficiency oftheengine is (4) 30% (n = 0.3). Calculate the loss of heat (Q, =?) per hour (t = 3600 sec). t =1hr = 3600 sec, n=30%=03, , =? Hint- P =50kW = 50,000 W, w P=— t S&g t ; W = P x t = W = 50,000 x 3600 = W = (180 x 10°] w

(ii)

 

 

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=0="53 =[600x 10°] =, W=0,-0, 20:=0-W =  Q, = 600 x  10° — 180 x10° =0Q,=420x10°] 1995 (420 x 10° J) In  a n Isobaric process when 2000] (AQ) of heat energy is supplied to a gas in a cylinder, the piston moves through 0.1 m (A x ) under a constant pressure of 2 x 1.01 x 10°N/m? (P). If the area of the pis ton is 5 x 1072 m? (4). Calculate the work done (AW =?) and the increase in internal (4) energy (AU =?) of the system. A=5x10"72m? Ax=01m, P=2x1.01x10° N/m? Hint- AQ =2000j, AU =? AW =?, AW = PAV @ = AW =P [(A)(Ax)] = > AW =2x101x10°x5x1072x0.1 = AW =[1010] AQ = AU + AW =  AU = AQ — AW =  AU = 2000 — 1010 = AU =990] 1996 (1010],990 J) Find the rms speed (9,,,,5) of nitrogen molecule at 27 °C (T = 300 K). Given the mass of 4) nitrogen molecule to be 4.67 x 102° kg (m), K = 1.38x 10723 J/K. T=27C%=27+273 =300K, m= 467x102 kg Hint- Opps = 02 =7, 3kT —_ — 92 = Opps = m

A=5x10"72m?

m

3 x 1.38x 10723 x 300 =  ms = 4.67 x 10-26

1.242 x 10-20 =  ms = 4.67 x 10°29 =  Opms = V265952.9 =  Oyms = [515.70 m/s 1996 (515.70 m/s) (9 Ql A glass flask is filled to the mark with 60 cm? (V,) of mercury at 20 °C (T, = 293 K). If theflask and its contents are heated to 40 °C (T, = 313 K), how much mercury will be above the mark? @ (Cass = 9 x 1076/°C and Bnerury = 182 x 1076/0C) Ty =  20 C* =20 +273 = 293K, ; = Vowereury = 600 m ®= 602107°m?, cm? > m3 + by 1x10° Hints Vou; = = T, =  40C° =40 +273 = 313K, AV = Vay orig Vegas = AV = Viyercury = Vigiass =  AV = Vouereury (1+ Butercury AT) = Voges, (1+ Botass AT) = r (UF Batass AT) Vopr, = , (1+ Butercury AT) = Voyeur ( =  AV = Vou, ( =   AV = Vou rey [1 + Batercury AT = (1+ Boras AT] Ve= =   AV = Vorereury [E+ Batercury AT = 1 = Batass AT] uteu Ve = = AV = Vouyer [Butercury AT = Botass AT]

Ql

(9

= Vopr, =

 

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=  AV = Vopr cury[(Batercury — Batass JAT]
y = 3%iass (Tz = T1)]
=  AV = Vo,ercury [(Brercury =
Vrr( = AV = 60x 10°[(182 x 107° = 3x 9x 107° )(313 — 293)]

Q7

 

m3 — cm? x by 1x 10°

Q.2 (d)

=  AV = 60x   107° [(182 x107° — 27 x107° )(20)]
=  AV = 60x  107°[(155 x107° )(20)]
= > AV =60x10"°x3.1x10"
=  AV = 0.186 x 10° m?*
m®
AV = 0.186 cm®
m3 — cm? x by 1x 10°
F in al mark of mercury in flask = 60 + 0.186 = 60.186 cm*
1997
(AV = 0.186 cm?, final mark of mercury in flask = 60 + 0.186= 60.186 cm®)
A Carnot engine whose low temperature reservoir is at 200 K (T,) has an efficiency of 50 % (1,).
Itis desired to increase this to 75% (n,). By how many degrees must the temperature decreased
(4)
(AT =T, — T) =7), if higher temperature (T,) of the reservoir remains constant?
AT =T, — T/ =?
1, = 75% = 0.75,
7, =50%=0.5,
Hint- T, =200K,

2
075-12 =075=1—-— 200
200

1
= 0 75 - 1 =
45

Tf
200

= T] = 400x025

= |T/ = 100K

AT=T,- T]
AT = 200 — 100
AT = 100K
1997
(100K)
Find the root-me a n square speed (¥,,,,) of a hydrogen molecule at 7 °C (T = 280 K). Take the
mass of a hydrogen molecule to be 3.32 x 1072” kg (m) and
Boltzmann's constant = 1.38 x 10723 J /K.
Hint- SameasQ.S5
(1868.6 m/s)

 

 

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540  calorie s (AQ = 2268] * 1 calorie= 4.2 ]) of he at is required to vaporize 1 gm of water at 100°C (T = 373 K). Determin e the entropy change (AS =?) involved in vaporizing 5 gm of water. (5) AS =? T =100C°=100+ 273 =373K, Hint:- AQ = 540 calorie = 2268],

Q2

QW

(9

T =100C°=100+ 273 =373K,

calorie & gt; | x by 4.2

Hint:- AQ = 540 calorie = 2268],

=  AS = 6.08//K (for 1 gm) AS = 6.08x5 AS =30.4]/K (for 5 gm) (304]/K) Find the efficiency (n =?) of a Carnot engine working between 100°C (T, = 373 K) and 500C (T, = 323K). T, =  50 C° =50 + 273 = 323K, Hint- T, =100C°=100+ 273 =373K,

T, =  50 C° =50 + 273 = 323K,

1999 (135%) C a lcul at e the volume (V =?) occupied by a gram-mole (n = 1) of a gas at 10 °C (T = 283 K) and (4) pressure (P) of one atmosphere. (Given:- R = 8.314] /mol.K,One atmosphere = 1.01x10° N /m?) Hint- V n= 1mole, T=10C"=10+273 =283K, P=1atm=101x10°N/m? R= 8314] /mol.K PV =nRT nRT P 1x8314x283 = & gt; V = 1.01x10° =  | V =10.0233m? = 23.3 litres|| 1L = 1000 m3 2000 (0.0233 m? OR 23.3 litres) In  a n Isobaric process when 2000 J (AQ) of heat energy is supplied to a gas in a cylinder the piston moves through 0.5 m (Ax) under a constant pressure of 1.01 x 10°N/m?(P). If the area of the  p is ton is 2 x 1072 m? (A). Calculate the work done (AW =?) and the increase in internal (5) energy (AU =?) of the system. Hint:- Same as Q.4 2000 (1010],990]) If one mole (n = 1) of a mono at omic gas is heated at constant pressure from -30 °C (Ty = 243 K) to 20 °C (T, = 293 K), find the change in its internal energy (AU =?) and the work (4) done (AW =?) during the process. o J (Cp = 20.8] /mol.K, C, = 12.5 J /mol.K) Hint- n= 1mole, T, =20C°=20+273 =293K, J T, =   —30 C° =—30 + 273 = 243K, = m AW =? AU =2, =  C, = 12.5] /mol.K, Cp = 20.8] /mol.K, For Isobaric process : — (i) “PAV = n R AT & a m p; Cp — Cy =R AW = PAV =nRAT = n(Cp — Cy)AT AW = 1x (20.8 — 125) x (T, = Ty) = AW =83x (293 — 243) = AW =83x 50 = [AW =415]

2000

(i)

“PAV = n R AT & a m p; Cp — Cy =R

 

 

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@ c= =  AQ, = nCpAT =  AQ, = 1x20.8x 50 =  [AQ, = 1040] AQ, = AW + AU =  AU = AQ, — AW =  AU = 1040 — 415 =[AU=625] (625],415])

2001

An ideal heat engine operates in Carnot’s cycle between temperatures 227 °C (T, = 500 K) and 127 9C (T, = 400 K) and it absorbs 600 J (Q,) of heat energy; find the:- (4) (ii) Work done per cycle (W =?) Efficiency of the engine (n =?) (i) T, =  127 C° =127 + 273 = 400 K, ( Hint- T, =227C°=227+273 =500K, Q, = 600], w=? ® n (it)

(4)

(it)

n

(iD)

n Ta w 0.2 =202=— 600 600 = > W =0.2x 600 = |W =120]

2001 (120,20 %) Find the Root Me a n Square velocity (9,,,s =?) of a hydrogen gas molecule at 100 °C (T = 373K). Take the mass (m) of the hydrogen molecule to be 3.32 x 10727 kg and K = 1.38 x 107% J /K. 4)

4) Hint- SameasQ.5 2002 P/M (2156.7 m/s) When 2000] (AQ) of heat energy is supplied to a gas in a cylinder at constant pressure of 1.01 x 10°N/m? (P), the piston of area of cross-section 2 x 1072 m? (A) moves through 0.5 m (Ax); calculate the work done (AW =?) and the increase in internal energy (AU =?) of the (4) system. Hint:- Same as Q4 2002 P/M (1010J,990)) A  heat engine performs 1,000 J] (W) of work and at the same time rejects 4,000 J (Q,) ofheat energy to  the cold reservoir. What is the efficiency (n =?) of the heat engine? If the difference of temperature between the sink and the source of th is engine is 75 0C (AT = T, — T,=75 C° =75 K), find the temperature of its source (T; =?)? Hint- W = 1000], T, =? AT =T,—T,= 75C°=75K, 7 , =4000], W = Q0-Q =Q ® SQ =W+Q, =  Q= Q, = 1000 + 4000 = [Q, = 5000]

 

 

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2002 P/E (20 %, 375 K) A cyl in der of di a meter 1.00 cm (D/) at 30 °C is to be slide intoa hole ina steel plate. The hole has a diameter of 0.99970 cm (D,) at 30 °C (T; =303 K). To what temperature (T, =?) must the plate 4) be heated? (ape; =1.1x 107° /°C) Hint- D/=1cm, T, =? T, =30C° =30+273 =303K, D, = 0.99970 cm, D/ = Dg (1+ aAT) = > —=1+a(T,— Ty) Do —_—  = 5 - = g9970 = 1+ [1121079 x (T, 303)] = 1.00030 = 1+ 1.1x 1075 x (T, — 303) = 1 .00030 — 1 =1.1x 1075 x (T, — 303) =  0.0003 = 1.1.x 1075 x (T, — 303) 00003 303 527.27 =T, 303 30822727 =T- Sirios iiri =T, =303 + 27.27 =|T, =330.27K 2002 P/E (330.27K) K The low temper a ture reservoir of a Carnot engine is at 5°C (T, = 278 K) and has an efficiency of 40 % (n,). Itis desired to increase its efficiency to 50% (n;). By what degrees should the (4) temperature of the hot reservoir be increased (AT = iv -T,=7)? 7, = 50% = 0.50, 1, =40%=04, Hint- T,=5C°=5+273=278K, AT= T/-T, =? Tz i z =1-2 Th iA 278 T,

T, =30C° =30+273 =303K,

303 527.27 =T, 303 30822727 =T-

(4)

 

 

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T, = 4633K

AT = 556 — 463.3 AT =92.7K

2003 P/M V;) conta ins 2 kg (m,) of air at a pressure of ?) would have  to be forced in to the tank to (4) ng no change in temperature? P,=18atm, 15 at m,

A 1 in H

2003 P/M 1200] (AQ) of heat  energy is supplied to the system at constant pressure. The internal energy of the system is increased by 750] (AU)and the volume by 4.5 m3 (AV). Find the work done (4) (AW  =?) against  the pist on and the pressure (P =?)onthe piston. Hint- SameasQ.1 2003 P/E (450J,100 N/m?) A Carnot engine whose low temperature reservoir is at 5°C (T, = 278 K) has an efficiency of 40 % (ny). Itis desired to increase this efficiency to 50 % (n,). By what degree should the (4) temperature of high-temperature reservoir be increased (AT = T/ -T,=2)? Hint- SameasQ.19 2003 P/E (927K) Calculate the density (p =?) of hydrogen gas, considering it to be an ideal gas, when the root mean square velocity of hydrogen molecules are 1850 m/s (J,,s) at 0°C (T = 273 K) and (4) 1 atmosphere pressure (P). (1 atmospheric pressure = 1.01 x 10°N/m?) T = 0C"=0+273 = 273K, , =1850m/s, Hint p=? 0, 0,

(4)

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P=1atm=1.01x10°N/m? pms = OF = 1850 =  (/97)? = (1850)? =  92 = 3422500 m?/s* 15 15 P= 3p? 3p

=2p= 5 3x1.01x10° =Sp= 3422500 =  |p = 0.08853 kg/m* 2004 (0.08853 kg/m® OR 88.53 x 10° kg/m?®) The low temper a t ure reservoir of a Carnot engine is at -3 °C (T, = 270 K) and has an efficiency of 40 %  (n,). Itis desired to increase the efficiency to 50 %(n,). By how many degrees should the (4) temperature of the hot reservoir be increased (AT = T/ -T, =7)? Hint- Sameas Q.19 2004 (90K) A br a ss ring of 20 cm (D,) diameter is to be mounted on toa metal rod of 20.02 cm (D”) diameter at 20 °C (T, = 293 K). To what temperature (T, =?) should the ring be heated? 4) (@prass = 1.9 107%/0C) Hint:- Sameas Q.18 2005 (345.63K) A 10 0 gm (m¢,) copper block is heated in boiling water (T,, = 373 K) for 10 minutes and then it is dropped in to 150 gm (M at er) Of water at 30 °C (Teqiorimeter = Twater = 303K) ina 200 gm (Mcqrorimeter) Calorimeter. If the temperature of water is raised to 33.6 °C (Tina = 306.6 K), determine  the specific heat (Ccatorimerer =?) Of the material of the (4) calorimeter? (For copper C,, = 386 J/Kg.K) Hint:- mg, = 100 gm = 0.1 kg, g —» Kg + by 1000 Myater = 150 gm = 0.15 kg, Te =  100 C° =100 +273 = 373K, Meatorimeter = 200 gm = 0.2 kg, Teatorimeter = Twater =30C°=30+273 =303K, Cuater = 4200 /Kg.K , Ceatorimeter =? Trina =  33.6 C° =33.6 +273 = 306.6 K, Heat  lost by Copper = Heat gained by Calorimer + Heatgainedby Water (Mey. Cu AT = Me a torimeter- Ceatorimer-AT + Mater. Cwater- A Te eaa ere ar m a (mew Cou Trina = Tew) = Meatorimeter- Ceatorimer- (Trina t = Teatorimeter) + Mwater- Cwater- (Trina — Twater) ( = )  x 0.1x 386 x  (306.6   — 373) = 0.2 x Ceaiorimer X (306.6— 303) + 0.15 x 4200 x(306.6—303) —38.6 x (66.4) = 0.2 x Ceqorimer X (3.6) + 630 x(3.6) 2563.04 = 0.72 X Coqrorimer + 2268 2563.04 — 2268 = 0.72 X Ceqiorimer 295.04 = 0.72 x Ceaiorimer 295.04 Ceatorimer =  — g =>— Ceatorimer = 409.8] /Kg.K 2005 (409.8]/KgK)

(4)

4)

g —» Kg + by 1000

2005 (409.8]/KgK) Find the change in volume (AV =?) of a brass sphere of 0.6 m (d) diameter when it is heated 4) from 30 °C  (T,  = 303K) t0 100 °C(T,= 373 K). (@pyrass = 19 x 1076/0C) T, =  100 C° =100 +273 = 373K, Hint- T, =30C°=30+273 =303K, AV =? d=06m, AV = BVAT AV = (3a) [sr] ar a 4 d\? )(3) 4 d\ a = @a)lzn(3) ke @-T)

 

 

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0.6 4 4 0.6 AV =( 3 x 19x107%) x3 3.142 x =? x (373 — 303) 12.568 AV =57x1070x x (0.3)? x70 —— 7.163 x107* — x 0.027 x70 AV = — 3 — AV =2.387x10"* x 1.89 AV =4.51x10"* m® 2006 (4.51x107*m%) A Celsius thermometer in a laboratory reads the surrounding temperature as 3 0°C (T;); what is (4) the  temperature in the Fahrenheit scale  (T, =?) and in absolute scale(T, =?)? Ty =? Tp=?, Hint- T.=30C°, Tp  =18Tc + 32 = Tp = 18x30 + 32 Tp =54+32 Ty = 86 F° Tq = To + 273 Ty = 30 + 273 Ty =303K 2006 (86 °F, 303K) Calculate the densit y (p =?) of hydrogen gas, considering it to be an ideal gas, when root mean squ are velocity (9,,,s) of hydrogen molecules are 1850 m/s (J,,s) at 0°C (T = 273 K) and (4) 1 atmosphere pressure (P). (1 atmospheric pressure = 1.01 x 10°N/m?) Hint:- Same as Q.23 2007 (0.08853 kg/m? OR 88.53 x 10-3 kg/m?) A heat engine performs work at the rate of 50 kW (P = 50,000 W). The efficiency oftheengine is 3) 30% (n = 0.3). Calculate the loss of heat (Q, =?) per hour (¢t = 3600 sec). Hint:- SameasQ.3 2007 (420 x 10°)) A steel bar is 10 m (L) in length at -2.5 °C (T; = 270.5 K). What will be the change in its length7/C (4) (AL =?) when it is at 25 °C (T, = 298 K)? (Bsteer = 3.3 x 1075 /0C) Hint- L=10m, T,=-25C°=-25+273=2705K, T,=25C"=25+273=298K,

AL = ar, -T) 33x107° L="""" x 10x (298 — 270.5) AL=11x10"5x10x 27.5 AL=3.025x10"°m 2008 (3.025x 105m) A Carnot engine performs 2,000 J (W)of work and rejects 4,000 J (Q,) of heat to the sink. If the difference of temperature between the source and the sink is 85 0C (AT = T, — T,=85 C° =85 K), find  the temperatures of the source (T; =?) and the sink (T, =7). T, Hint- W=2000], AT=T,-T,=85C°=85K, Q, =4000j, W=0,-20, 20=W+Q, =  Q; = 2000 + 4000 = [Q, = 6000)

AT=T,-T,=85C°=85K,

Q, =4000j,

T,

= 1 - 2 1[o [o)

 

 

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T, =257.57K 0 T, T, ST, ==2xT, T,  4800 257.57 = >T,= — — | > 5000 *

= T, = 0.67 x 257.57 =|T, =172.57K ( 2 A de ( A Hi

2008 tmospheres (P;). Overnight the tank (V; = V,) to 950 atmospheres (P,). Calculate the mass 4 Am = my, —m,| =? 5 0 atm, V,=V,,

5 0 atm, V,=V,,

. , Am =m; —my Am = [17.40 — 22| Am =4.6 gm 2009 (4.6 gm) In  a n Isobaric process when 2000 J (AQ) of heat energy is supplied to a gas in a cylinder, the piston of area 2 x 1072 m? (A) moves through 40 cm (Ax = 0.4 m) under a pressure of (4) 1.01 x 10°N/m? (P). Calculate the increase in internal energy (AU =?) of the system. Hint:- same as Q4 2009 192]) A heat eng in e performing 400] (W) of work in each cycle has an efficiency of 25% (17). How (4) much heat is absorbed (Q,  =?) and rejected (Q,=?) in each cycle? Hint- W = 400], n=125%= 0.25, Q, =? Q, =2, w

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= [Q; = 1600] W=0-20; =20=0-W =  Q, = 1600 — 400 = [Q, = 1200) 2010 (1600],1200]) (vi) Q.2 A200 g (Meta = 0.2 kg) piece of metal is heated to 150 °C (Type¢q = 423 K) and then dropped in to an aluminum calorimeter of mass 500 g (M atorimeter = 0.5 kg), containing 500 g (Myyqter = 0.5 kg) of wat er initially at 25 °C (Taiorimeter = Twater = 298 K). Find the final (4) equilibrium temperature (Tino =?) of the system. (Cmetar = 128.100 J/Kg. K, Ceatorimeter = 903 ]/Kg. K, Carer = 4200] /Kg.K) 9 = Kg = by 1000 Myater = 500 gm = 0.5 kg, Hint- Mypeeq = 200 gm = 0.2 kg, Teta =  150 C° =150 +273 = 423 K, Meatorimeter = 500 gm = 0.5 kg, Trina =? Teatorimeter = Twater = 25 C° =25 +273 = 298K, Heat  lost by Metal = Heat gained by Calorimer + Heatgainedby Water (Mmetar- Cetat- AT = Mcatorimeter-Ceatorimer-AT + Muater- Cater. ATmcoectew tue (Mmetar- Cmetat- Trinat — Tmetat) = Mcatorimeter- Ccatorimer (Trinat = Teatorimeter) + Mwater- Cwater (Trinat— Twater) ( = ) x0.2x 128.1 x (Tjinq — 423) = 0.5 x 903 x (Tying — 298) + 0.5x 4200 x(Tying—298) (-) 25.62 x (Tina — 423) = 451.5 x (Trina — 298) + 2100 x (Tyne —298) ( = ) 25.62 x (Tyinar — 423) = (Trina — 298)(451.5 + 2100) (-) 25.62 x (Tyinq — 423) = (Trina — 298) x 2551.5 2551.5 55 OO) (Tpinat = 428) = (Tyna = 2 98) X55 9 (i a 55 (Tyne s 423) = (Tyne — 298) 99.60 9 ()(Trinar — 423) = 99.60 x Tina — 298 x99.60 Ty in at + 423 = 99.60 x Tying —29680.8 Ty in at — 99.60 X Tying = —29680.8 — 423 —100.60 x Tying = —30103.8 30103.8 Trinat = 7 760.60 Trina = 299.24 K

Q.2

Q36

(vi)

9 = Kg = by 1000

2011 (299.24K) A Carnot engine whose low temperature reservoir is at 200 K (T,) has an efficiency of 50 % (n,). Itis desired to increase this to 75 % (n,). By how many degrees must the temperature of low temperature reservoir be decreased (AT =T, — T) =7), if the temperature (T;) of the higher (4) temperature reservoir remains constant? Hint:- Same as Q.7 2012 (100K) The difference of temperature between a hot and a cold body is 120 °C (AT). If the heat engine is efficient 30 % (n), find  the  temperature of the hot (7; =?) and the cold (T, T, =?, Hint:- AT =120°C =120K, n=30%=0.3,

 

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AT = T, T,
= 120 = 400 — T,
= T, = 400 — 120
=[T, =280K
(400K, 280 K)

 and the