1. This document discusses concepts of heat and temperature including specific heat capacity, latent heat of fusion and vaporization, and using the concepts of heat transfer and phase changes to solve calculation problems. 2. Several examples are provided to demonstrate calculating heat transfer involved in temperature changes of substances using their specific heat capacities as well as phase changes using latent heat values. 3. Formulas used include Q=mcΔT to calculate heat transfer due to temperature change based on specific heat capacity c, and Q=mL to calculate heat of phase change based on latent heat L.

1.    SchoolDD.com
10
www.schoolDD.com

2. 10
10.1 F
F ˈ ˂
F ˈ F F (Q) F ˈ (J)
- F 1 (= 4.186 J ) F F 1
1° C
10.1.1
ˈ F F
F F F F F
F F F
10.2.1 F F
F F ˈ F F F F F
F F F F F
F F F
1. (° C)
2. (K) ˈ F F FF F F F ( ˈ F
SI)
K °C
373.16 100
F F K = C + 273
273.16 0
F
C=
۱ ‫܀‬ ۴ି૜૛ R= F
= =
૞ ૝ ૢ F= F
K=
– F ˆ F www.schoolDD.com 1

3. 10.1.2 F
1. F (C) F F
ˈ F F ˈ F (J/K)
m
〉 ∆T ઢ‫ۿ‬
C=
∆Q ઢ‫܂‬
2. F (c ) F F 1 F
ˈ F F ˈ F (J/kg K)
ઢ‫ۿ‬
〉 ∆T
m
c=
∆Q
‫ܕ‬ઢ‫܂‬
F FF F F m ΔT F F
Q= F (J)
m= (kg)
Q = mc∆T
c= F (J/kg K)
ΔT = (K ˚C)
- F (C) F F F C F C F
- F (c) F F ˈ
F c (J/kg K)
4186
(-10˚C 0˚C) 2100
900
450
390
140
130
– F ˆ F www.schoolDD.com 2

4. F 1 F F 100 25˚C ( c = 390
J/kg K)
. ˈ 45 ˚C
. ˈ 15 ˚C
. F F F F F
Q = mc∆T
= 0.10x390x(45-25)
‫ ׵‬Q = 780 J Ans
. F F F
Q = mc∆T
= 0.10x390x(25-15)
‫ ׵‬Q = 390 J Ans
10.1.4 F
F F F F
F F F
F
s
- ˈ F
F F F F F F F
10.1.5 F
3
F F F F F
F F
F F F F F
F F F F F
F F F F F F F F F
F F ˂
– F ˆ F www.schoolDD.com 3

5. F F
F
F
F
F
F F m F F F
Q= F (J)
Q = mL m= (kg)
L= F (J/kg)
F F
Lm (J/kg) ˈ Lv (J/kg)
333x103 2256x103
58.6x103 452x103
5.23x103 20.9x103
F 2 F F 200 -10˚ C ˈ
F ˈ F 190 ( c = 2100 J/kg K, c =
3 3
4186 J/kg K, Lm = 333x10 J/kg, Lv = 2256x10 J/kg )
“ F F ˈ
F F F F ˇ F F ”
Q1 Q2 Q3 Q4
200 g 200 g 200 g 200 g 190 g + 10 g
-10° C 0° C 0° C 100° C 100° C
– F ˆ F www.schoolDD.com 4

6. F F 200 g -10° C ˈ 0° C
Q1 = mc∆T
= 0.2x2100x(0-(-10))
Q1 = 4200 J
F F 200 g 0° C ˈ 0° C
Q2 = mL
= 0.2x333x103
Q2 = 66600 J
F F 200 g 0° C ˈ 100° C
Q3 = mc∆T
= 0.2x4186x(100-0)
Q3 = 83720 J
F F 10 g 100° C ˈ 10 g 100° C
Q4 = mL
= 0.01x2256x103
Q4 = 22560 J
‫׵‬ F Q = Q1 + Q2 + Q3 + Q4
= 4200 + 66600 + 83720 + 22560
Q = 177080 J Ans
F 3
100 100˚ C F F F ˈ
90˚ C (c = 4186 J/kg K, Lv = 2256x103 J/kg)
“ F
”
Q1 Q2
100 g 100 g 100 g
100° C 100° C 90° C
F 100 g 100° C F ˈ 100 g 100° C
Q1 = mLv
= 0.1x2256x103 (L =L F )
Q1 = 225600 J
F 100 g 100° C ˈ 100 g 90° C
Q2 = mc∆T
= 0.1x4186x(100-90)
Q2 = 4186 J
‫׵‬ F F Q = Q1 + Q2
= 225600 + 4186
Q = 229786 J Ans
– F ˆ F www.schoolDD.com 5

7. F 4
F F 1 20 ˚C F ˈ -10 ˚C F
F F (c = 2100 J/kg K, c = 4186 J/kg K, Lm = 333x103 J/kg)
“ F
”
Q1 Q2 Q3
1 kg 1 kg 1 kg 1 kg
20° C 0° C
0° C -10° C
F 1 kg 20° C ˈ 1 kg 0° C
Q1 = mc∆T
= 1.0x4186x(20-0)
Q1 = 83720 J
F 1 kg 0° C ˈ 1 kg 0° C
Q2 = mLm
= 1.0x333x103 (L =L )
Q2 = 333000 J
F 1 kg 0° C ˈ 1 kg -10° C
Q3 = mc∆T
= 1.0x2100x(0-(-10))
Q3 = 21000 J
‫׵‬ F F Q = Q1 + Q2 + Q3
= 83720 + 333000 + 21000
Q = 437720 J Ans
10.1.6 F
F F F F
F F F F
F F F F
F = F F
Q =Q Q = Q
F 5
F 200 FF 300˚ C F 1 20
˚C F F ˈ F ( c = 500 J/kg K, c = 4200 J/kg )
– F ˆ F www.schoolDD.com 6

8. 0.2 kg 0.2 kg
Q Q
300° C t° C t ≤ 100° C F ˈ
t = 300° C F
1 kg 1 kg
100° C ˈ F
Q 20° C Q t° C
Q =Q
(mc∆T) = (mc∆T)
F F ˈ t° C 20 < t ≤100
F F 0.2x500x(300- t) = 1.0x4200x(t-20)
30000 - 100 t = 4200t – 84000
114000 = 4300t
‫ ׵‬t = 26.5° C Ans ( )
F 6
F 5 F F ˈ 2 ˈ 0.5
F ˈ F
2 kg 2 kg
Q Q
300° C t° C t ≤ 100° C F ˈ
t = 300° C F
0.5 kg 0.5 kg
Q 100° C ˈ F
Q 20° C t° C
Q =Q
(mc∆T) = (mc∆T)
F F ˈ t° C 20 < t ≤100
F F 2.0x500x(300- t) = 0.5x4200x(t-20)
30000 - 100 t = 2100t – 42000
342000 = 3100t
‫ ׵‬t = 110.3° C > 100° C F ˈ F ˈ
F ˈ 100° C F F
2 kg 2 kg
Q Q 100° C
300° C
0.5 kg 0.5 kg +
Q 20° C Q 1 100° C Q 2
100° C
– F ˆ F www.schoolDD.com 7

9. ˈ F F
Q =Q
(mc∆T) = (mc∆T) 1 + (mL) 2
F F 2.0x500x(300-100) = 0.5x4200x(100-20) + mx2256x103
32000 = 2256x103
‫ ׵‬m = 0.014 kg
ˈ 0.014 kg 100° C F F 0.5 0.014 = 0.486 kg 100° C Ans
F 7
˄ 1 2 20˚ C F FF 1
200 ˚C F ˈ F ( c = 1000 J/kg K, c
= 4200 J/kg c = 400 J/kg K )
1 kg 1 kg
Q 200° C
Q t° C
˄ 1 kg ˄ 1 kg
t° C 20 < t ≤ 100° C F ˈ
20° C Q ˁ
Q
2 kg 2 kg
20° C Q t° C
Q =Q
(mc∆T) = (mc∆T) ˄ + (mc∆T)
F F ˈ t° C 20 < t ≤100
F F 1.0x400x(200- t) = 1.0x1000x(t-20) + 2.0x4200x(t-20)
80000 - 400 t = 1000t – 20000 + 8400 t - 168000
268000 = 9800t
‫ ׵‬t = 27.35° C Ans ( )
F 8
F 200 1 30 ˚C F 100
0 ˚C F ( c F = 800 J/kg K, c = 4200 J/kg K,
L = 333x103 J/kg )
– F ˆ F www.schoolDD.com 8

10. 1 kg 1 kg
30° C Q t° C
Q
F 0.2 kg F 0.2 kg 0 < t < 30° C
30° C Q F
t° C
0.1 kg 0.1 kg
0.1 kg
Q Q 0° C Q t° C
0° C
Q =Q
(mc∆T) + (mc∆T) F = (mL) + (mc∆T)
F F ˈ t° C 0 < t < 30
F F 1.0x4200x(30- t) + 0.2x800x(30- t) = 0.1x333x103 + 0.1x4200x(t-0)
4360x(30- t) = 33300 + 420t
97500 = 4780t
‫ ׵‬t = 20.4° C Ans ( )
F 9
F 1 100 ˚C F 200 F
ˈ F ( c = 500 J/kg K, c = 4200 J/kg K, L = 333x103 J/kg )
1 kg 1 kg
Q 100° C Q t° C
0 < t < 100° C
0.2 kg 0.2 kg
0.2 kg
Q Q 0° C
Q t° C
0° C
Q =Q
(mc∆T) = (mL) + (mc∆T)
F F ˈ t° C 0 < t < 100
3
F F 1.0x500x(100- t) = 0.2x333x10 + 0.2x4200x(t-0)
50000 - 500t = 66600 + 840t
-16600 = 1340t
– F ˆ F www.schoolDD.com 9

11. ‫ ׵‬t = -12.39° C F ˈ F F ˈ 0° C
ˈ F
F F
1 kg 1 kg
Q 100° C Q 0° C
0.2 kg +
Q 0° C
Q 0° C
Q =Q
(mc∆T) = (mL)
F F 1.0x500x(100- 0) = mx333x103
50000 = 333x103m
‫ ׵‬m = 0.15 kg = 150 g < 200 g OK
F F ˈ 0° C 0.15 kg Ans
10.1.7 F F
3
1. F
- F F F ˈ F F F
F F F
2. F
- F F ˈ F F F
F F F F F F F
FF F F F
3. F F
- F F F F F F F
F ˂
– F ˆ F www.schoolDD.com 10

12. 10.2 F
- F F F F F F
F ˈ F
10.2.1 F
F ʽ F F
P
૚ T
V ‫ן‬
‫۾‬ PV = F 1
V
10.2.2 F
F ʽ F F
V
V ‫ן‬T ୚
P
୘
= F T
10.2.3 F
F F F
୘
‫ן‬
୔
V
୔୚
= F
୘
‫۾‬૚ ‫܄‬૚ ‫۾‬૛ ‫܄‬૛
F F F
F = 1 2 (T F )
‫܂‬૚ ‫܂‬૛
F F P, V F F = F F F F T
F ˈ K F
– F ˆ F www.schoolDD.com 11

13. F
‫ן‬n
୔୚ ୫
F ୘
n= F =
୑
1
‫۾‬૚ ‫܄‬૚ ‫۾‬૛ ‫܄‬૛
= F F F 1 2
F ‫ܖ‬૚ ‫܂‬૚ ‫ܖ‬૛ ‫܂‬૛ F (T F )
ˈ F
PV = nRT R= F F = 8.31 J/mol K
. F
୒
n= ‫ۼ‬
‫ۯ‬
= 6.02 x 1023
F 1 F 6.02x1023
୒
F PV =
‫ۯۼ‬
RT
ୖ ଼.ଷଵ
F F KB =
‫ۯۼ‬
= F F F =
଺.଴ଶ୶ଵ଴మయ
= 1.38x10-23 J/K
F PV = NKB T
PV = nRT PV = NKB T F F F F
F ˆ F F F
1. F PV = nRT
ଵ
2. F F F ‫ ן‬T, P ‫ ן‬n, V ‫ ן‬T, V ‫ ן‬n ‫ן‬ F
୚
P P
3. F F F F F 1 2 F F F
F F
୔భ ୚భ ୔మ ୚మ
= R = R
୬భ ୘భ ୬మ ୘మ
୔భ ୚భ ୔మ ୚మ
୬భ ୘భ
=
୬మ ୘మ
(P F ˈ P F )
F F
– F ˆ F www.schoolDD.com 12

14. F 10
F 1 F 2 x 105 Pa 7˚ C F F F F
2 F F ˈ F
V 1 = 1 m3 V 2 = 2 m3
P1 = 2x105 Pa P2 = P 1
T1 = 273+7 = 280 K T2 = ?
n1 n2 = n1
F PV = nRT
F F 2
୔భ ୚భ ୔మ ୚మ
=
୬భ ୘భ ୬మ ୘మ
P n
୚భ ୚మ
F ୘భ
=
୘మ
ଵ ଶ
=
ଶ଼଴ ୘మ
‫ ׵‬T2 = 560 K 560 273 = 287° C Ans
F 11
F F 105 Pa F 27˚ C ˈ -3˚ C ˈ
F
V1 V2 = V1
P1 = 2x105 Pa P2 = ?
T1 = 273+27= 300 K T2 = 273+(-3) = 270 K
n1 n2 = n1
F PV = nRT
F F 2
୔భ ୚భ ୔మ ୚మ
=
୬భ ୘భ ୬మ ୘మ
V n
– F ˆ F www.schoolDD.com 13

15. ୔భ ୔మ
F ୘భ
=
୘మ
ଵ଴ఱ ୔మ
=
ଷ଴଴ ଶ଻଴
‫ ׵‬P2 = 0.9x105 Pa Ans
F 12
F F F 17˚ C F 2 x 105 Pa F
F ˈ 27˚ C F ˈ F (
=105 Pa)
V1 V2 = V1
Pg1 = 2x105 Pa Pg2 = ?
T1 = 273+17= 290 K T2 = 273+27 = 300 K
n1 n2 = n1
F PV = nRT
F F 2
୔భ ୚భ ୔మ ୚మ
=
୬భ ୘భ ୬మ ୘మ
V n
୔భ ୔మ
F ୘భ
=
୘మ
ሺ୔ౝశ ౌబ ሻభ ሺ୔ౝశ ౌబ ሻమ
୘భ
=
୘మ
F P FF ˈ P F F
ଶ୶ଵ଴ఱ ାଵ଴ఱ ୔ౝమ ାଵ଴ఱ
=
ଶଽ଴ ଷ଴଴
‫ ׵‬Pg 2 = 2.09x105 Pa Ans
F 13
F 10 27˚ C 105 Pa F
F
– F ˆ F www.schoolDD.com 14

16. V = 10
P = 105 Pa
T = 273+27 = 300 K
n=?
m=?
F PV = nRT
F F F F
F F 105(10x10-3) = n (8.31)(300) 1 = 1000 cc = 1000 cm3 = 1000(10-6 m3) = 10-3 m3
‫ ׵‬n = 103/ 2493 = 0.4 Ans
୫
n= ୑ m = nM
= 0.4x2.0x10-3
‫ ׵‬m = 8.0x10-4 kg Ans
F 14
F 50 F 10 27˚ C
F F F F 5 17˚ C F F F
( R = 8.31 J/mol K, MN2 = 28 g/mol, 1 atm = 105 Pa )
V1 = 50x10-3 m3 V2 = 50x10-3 m3
P1 = 10 atm P2 = 5 atm
T1 = 273+27 = 300 K T2 = 273+17 = 290 K
n1 = ? n2 = ?
. n1 F F P1 V1 = n1RT1
10x105x50x10-3 = n1x8.31x300
n1 = 20
. n2 F F P2 V2 = n2RT2
5x10 x50x10-3 = n2x8.31x290
5
n2 = 10.4
‫׵‬ F F F n1 n2 = 20 10.4 = 9.6 9.6x28x10-3 = 0.27 kg Ans
– F ˆ F www.schoolDD.com 15

17. F 15
F 64 27˚ C 20 F F
F ( R = 8.31 J/mol K, MO2 = 28 g/mol )
m = 64 g
T = 273+27 = 300 K
V = 20x10-3 m3
P=?
P F F PV = nRT
଺ସ ୫
Px20x10-3 = ଷଶx8.31x300 n= ୑
‫ ׵‬P = 2.493x105 Pa Ans
F 16
F 1 1.3 0˚ C 1
13 10 27˚ C
V1 = 1 V2 = 10
m1 = 1.3 g m2 = 13 g
T1 = 273+0 = 273 K T2 = 273+27 = 300 K
P1 = 1 atm P2 = ?
F PV = nRT
F F 2
୔భ ୚భ ୔మ ୚మ
=
୬భ ୘భ ୬మ ୘మ
୫
F n= ୑
M
୔భ ୚భ ୔మ ୚మ
F ୫భ ୘భ
=
୫మ ୘మ
ଵ୶ଵ ୔మ ୶ଵ଴
=
ଵ.ଷ୶ଶ଻ଷ ଵଷ୶ଷ଴଴
‫ ׵‬P2 = 1.10 atm Ans
– F ˆ F www.schoolDD.com 16

18. F 17
F F F F
ˈ F
V1 V2 = V1
P1 P2 = ½ P 1
T1 T2 = T1
n1 n2 = ?
F PV = nRT
F F 2
୔భ ୚భ ୔మ ୚మ
=
୬భ ୘భ ୬మ ୘మ
F V T
୔భ ୔మ
F ୬భ
=
୬మ
భ
୔భ ୔
మ భ
=
୬భ ୬మ
‫ ׵‬n2 = ½ n1 Ans
F 18
F F 10 kg/m3 27˚ C 2 F F
F 17˚ C 1 F F F
V1 V2 = V1
T1 = 273+27 = 300 K T2 = 273+17 = 290 K
P1 = 2 atm P2 = 1 atm
ߩ1 = 10 kg/m3 ߩ2 = ?
F PV = nRT
F F 2
୔భ ୚భ ୔మ ୚మ
=
୬భ ୘భ ୬మ ୘మ
୫
F n= ୑
M
– F ˆ F www.schoolDD.com 17

19. ୔భ ୚భ ୔మ ୚మ
F ୫భ ୘భ
=
୫మ ୘మ
୔భ ୔మ
ౣభ = ౣమ
୘ ୘
౒భ భ ౒మ మ
୔భ ୔మ
=
ఘభ ୘భ ఘమ ୘మ
ଶ ଵ
=
ଵ଴୶ଷ଴଴ ఘమ ୶ଶଽ଴
‫ߩ ׵‬ଶ = 5.17 kg/m3 Ans
10.3 F F
10.3.1 F
F
1. F F F ˈ
F F F
2. F F ˈ F
3. F ˈ F F F
10.3.2 F ( < EK > )
F F < EK > ˈ F F
1 F ˈ (J)
< EK > = ૜ KB T
૛
F F N F N< EK >
ଷ
N< EK > = ଶ NKB T
ଷ
= ଶ PV ( PV = NKB T )
ଷ
= ଶ nRT ( PV = nRT )
- < EK > F F F F F
– F ˆ F www.schoolDD.com 18

20. 10.3.3 F
F F F vrms (vroot mean square )
F ˈ F (m/s)
૜‫܂܀‬
vrms = ට M= F 1 (kg)
‫ۻ‬
- vrms F T M
- vrms F F (v) F F F
F 19
F 20 27˚ C ( R = 8.31 J/mol K, MH2 = 2 g/mol )
. F F
. F F
. vrms F
m = 20 g
T = 273+27 = 300 K
. < EK > = ?
ଷ ୖ ଼.ଷଵ
< EK > = KB T KB = = = 1.38x10-23
ଶ ୒ఽ ଺.଴ଶ୶ଵ଴మయ
ଷ
= x 1.38x10-23x300
ଶ
‫׵‬ < EK > = 6.21x10-25 J Ans
. N< EK > = ?
. N N = n NA
୫ ଶ଴
= = x 6.02x1023
୑ ଶ
N = 6.02x1024
‫׵‬ N< EK > = 6.02x1024x 6.21x10-25 = 3.73 J Ans
– F ˆ F www.schoolDD.com 19

21. . vrms = ?
ଷୖ୘
vrms = ට ୑
ଷ୶଼.ଷଵ୶ଷ଴଴
= ට = √3739500
ଶ௫ଵ଴షయ
‫ ׵‬vrms = 1933.78 m/s Ans
F 20
F F F 3 10 2 Pa
n=3
V = 10 = 10x10-3 m3
P = 2 Pa
N<EK> = ?
<EK> = ?
. N N = n NA = 3x6.02x1023
ଷ ଷ
F F N< EK > = ଶ
NKBT = ଶ
PV (‫ ׶‬PV = NKBT)
ଷ -3
= x2x10x10
ଶ
‫׵‬ N< EK > = 3x10-2 J Ans
ଷ୶ଵ଴షమ ଷ୶ଵ଴షమ
F < EK > < EK > = ୒
=
ଷ୶଺.଴ଶ୶ଵ଴మయ
‫ < ׵‬EK > = 1.66x10-26 J Ans
F 21
F F 2x 1025 F F 105 Pa
F F
N/V = 2x1025 /m3
P = 1x105 Pa
<EK> = ?
ଷ
F < EK > < EK > =
ଶ
KBT
ଷ ଷ
F N< EK > = ଶ
NKBT =
ଶ
PV (‫ ׶‬PV = NKBT)
ଷ ୔୚
< EK > = ଶ ୒
ଷ ଵ୶ଵ଴ఱ
= ଶ x ଶ୶ଵ଴మఱ
‫ < ׵‬EK > = 7.5x10-21 J Ans
– F ˆ F www.schoolDD.com 20

22. F 22
F
. F F ˈ 2 F F ˈ F
. F F ˈ 2 F F ˈ F
.
T1 T2 = 2 T1
<EK1> <EK2> = ?
ଷ
< EK > = ଶ
KBT
F < EK > ‫ ן‬T
ழாేభ வ ୘భ ୘భ
= =
ழாేమ வ ୘మ ଶ୘భ
‫ < ׵‬EK2 > = 2< EK1 > Ans
.
V1 V2 = V1
P1 P2 = 2 P 1
N1 N2 = N1
<EK1> <EK2> = ?
ଷ
< EK > = ଶ
KBT
ଷ ଷ
N< EK > = ଶ
NKBT =
ଶ
PV
F < EK > ‫ ן‬P N V
ழாేభ வ ୔భ ୔భ
= =
ழாేమ வ ୔మ ଶ୔భ
‫ < ׵‬EK2 > = 2< EK1 > Ans
F 23
F A 1 7˚ C F B 2 27˚ C
F F ˈ F ( R = 8.31 J/mol K )
nA = 1 nB = 2
A TA = 273 + 7 = 280 K + B TB = 273 + 27 = 300 K
<EKA> <EKB>
– F ˆ F www.schoolDD.com 21

23. ଷ
N< EK > N< EK > = NKBT
ଶ
ଷ
= nRT
ଶ
ଷ ଷ
(N< EK >)A + (N< EK >)B = (ଶ nRT)A + (ଶ nRT)B
ଷ
= R(nATA + nBTB)
ଶ
ଷ
= x8.31(1x280 + 2x300)
ଶ
‫׵‬ F = 1.097x104 J Ans
F 24
F ˈ F F ( MH2 = 2 g/mol, MO2 =
32 g/mol)
H2
vrmsH2 O2
vrmsO2
ଷୖ୘
vrms = ට ୑
ଵ
F vrms ‫ן‬ T
√୑
୴౨ౣ౩ౄమ ୑ ଷଶ
୴౨ౣ౩ోమ
= ට୑ోమ = ට = 4
ౄమ ଶ
‫ ׵‬vrmsH2 = 4 vrmsO2 Ans
F 25
F ʽ F F ˈ 4 F
vrms ˈ F
ߩଵ ߩଶ = 4ߩଵ
vrms1 vrms2
ଷୖ୘
vrms = ට ୑
PV = nRT
୔୚
F RT = ୬
ଷ୔୚
vrms = ට ୬୑
– F ˆ F www.schoolDD.com 22

24. ଷ୔ ୬୑ ୫
= ටఘ = = ߩ
୚ ୚
ଵ
F vrms ‫ן‬ P
ඥఘ
୴౨ౣ౩భ ఘ ସఘ
= ටఘమ = ටఘభ = 2
୴౨ౣ౩మ భ భ
‫ ׵‬vrms2 = 2 vrms1 Ans
10.4
-
- F FF F F
F U F F
F F ˈ F F F
F F F F F F F
F F
U = N< EK >
ଷ
F FF U = ଶ
NKBT
ଷ
= nRT
ଶ
ଷ
= ଶ
PV
10.4.1 F 1 F F
F F F F
〉 ∆W
∆Q = ∆U + ∆W
〉 ∆U
〉 ∆Q
ΔW = FΔs
= PAΔs
= PΔV
– F ˆ F www.schoolDD.com 23

25. F F F +
∆Q
F -
+
∆U
-
+
∆W
F -
F 26
F F F 1000 J F F F 200 J F
F F F
∆W = ?
〉 ∆U = 200 J
〉 ∆Q = 1000 J
∆Q = ∆U + ∆W
1000 = 200 + ∆W
∆W = 800
‫׵‬ F F = 800 J ( ∆W ˈ + ) Ans
F 27
F 1024 F F F F 1˚ C F F F F F F
(KB = 1.38x10-23 )
N = 1024
∆T = 1
∆V = 0
〉 ∆Q = ?
∆Q = ∆U + ∆W
ଷ
= NKB∆T + P∆V
ଶ
ଷ
= x1024x1.38x10-23x1 + 0
ଶ
∆Q = 20.7
‫ ׵‬F F F F F ∆Q = 20.7 J Ans (∆Q ˈ +)
– F ˆ F www.schoolDD.com 24

26. F 28
100 N F 1 F 1K
F F F F 20 cm. ( R = 8.31 J/mol K )
F = 100 N
n=1
∆T = 1 K
0.20 ∆Q = ?
∆Q = ∆U + ∆W
ଷ
= ଶ
nR∆T + (-F∆s) ( F ˈ )
ଷ
= x1x8.31x1 – 100x0.20
ଶ
∆Q = -7.5 ( ˈ F )
‫׵‬ F ∆Q = 7.5 J Ans
F 29
F F F F F F F
P
∆W
∆U ∆V
〉 ∆Q = ?
∆Q = ∆U + ∆W
ଷ
= ଶ
NKB∆T + P∆V
ଷ
= P∆V + P∆V ( ‫ ׶‬PV = NKBT)
ଶ
ହ
∆Q = P∆V
ଶ
‫ ׵‬F F F F F ∆Q =
ହ
ଶ
P∆V J Ans
– F ˆ F www.schoolDD.com 25

27. F 30
F F F 20 27˚ C 105 Pa F
ˈ 57˚ C F F F F F F
V1 = 20
T1 = 273 + 27 = 300 K V2 = ?
∆W T2 = 273 + 57 = 330 K
P1 = 105 Pa
∆U P2 = P1
〉 ∆Q = ?
୔భ ୚భ ୔మ ୚మ
V2 F ୬భ ୘భ
=
୬మ ୘మ
P n
୚భ ୚మ
F ୘భ
=
୘మ
ଶ଴ ୚మ
=
ଷ଴଴ ଷଷ଴
V2 = 22
∆Q = ∆U + ∆W
ଷ
= NKB∆T + P∆V
ଶ
ଷ
=
ଶ
P∆V + P∆V ( ‫ ׶‬PV = NKBT)
ହ ହ
= P∆V = x10 x(22-2)x10-3
5
ଶ ଶ
∆Q = 500
‫ ׵‬F F F F F ∆Q = 500 J Ans
– F ˆ F www.schoolDD.com 26