1. SchoolDD.com provides information about heat transfer and calorimetry. It explains key concepts like specific heat capacity, latent heat of fusion and vaporization, and uses equations like Q=mcΔT. 2. Examples are given to calculate the heat transfer involved in changing temperatures of substances. Specific heat values are provided for various materials at different phases. 3. Phase changes from solid to liquid to gas are explained, along with the concept of latent heat absorbed or released without changing temperature during these phase transitions.

1. SchoolDD.com
10
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2. 10
10.1 F
F
F F F (Q) F (J)
- F 1 (= 4.186 J ) F F 1
1° C
10.1.1
F F
F F F F F
F F F
10.2.1 F F
F F F F F F F
F F F F F
F F F
1. (° C)
2. (K) F F FF F F F ( F
SI)
K °C
373.16 100
F F K = C + 273
273.16 0
F
C=
R= F
= =
F= F
K=
– F F www.schoolDD.com 1

3. 10.1.2 F
1. F (C) F F
F F F (J/K)
〉 ∆T
m C=
∆Q
2. F (c ) F F 1 F
F F F (J/kg K)
〉 ∆T
m
∆Q c=
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
F c (J/kg K)
4186
(-10˚C 0˚C) 2100
900
450
390
140
130
– F F www.schoolDD.com 2

4. F 1 F F 100 25˚C ( c = 390
J/kg K)
. 45 ˚C
. 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
- 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 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) Lv (J/kg)
333x103 2256x103
58.6x103 452x103
5.23x103 20.9x103
F 2 F F 200 -10˚ C
F F 190 ( c = 2100 J/kg K, c =
3 3
4186 J/kg K, Lm = 333x10 J/kg, Lv = 2256x10 J/kg )
“ F F
F F F F 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 F www.schoolDD.com 4

6. F F 200 g -10° C 0° C
Q1 = mc∆T
= 0.2x2100x(0-(-10))
Q1 = 4200 J
F F 200 g 0° C 0° C
Q2 = mL
= 0.2x333x103
Q2 = 66600 J
F F 200 g 0° C 100° C
Q3 = mc∆T
= 0.2x4186x(100-0)
Q3 = 83720 J
F F 10 g 100° C 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
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 100 g 100° C
Q1 = mLv
= 0.1x2256x103 (L =L F )
Q1 = 225600 J
F 100 g 100° C 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 F www.schoolDD.com 5

7. F 4
F F 1 20 ˚C F -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 1 kg 0° C
Q1 = mc∆T
= 1.0x4186x(20-0)
Q1 = 83720 J
F 1 kg 0° C 1 kg 0° C
Q2 = mLm
= 1.0x333x103 (L =L )
Q2 = 333000 J
F 1 kg 0° C 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 F ( c = 500 J/kg K, c = 4200 J/kg )
– F F www.schoolDD.com 6

8. 0.2 kg 0.2 kg
Q Q
300° C t° C t ≤ 100° C F
t = 300° C F
1 kg 1 kg
100° C F
Q 20° C Q t° C
Q =Q
(mc∆T) = (mc∆T)
F F 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 2 0.5
F F
2 kg 2 kg
Q Q
300° C t° C t ≤ 100° C F
t = 300° C F
0.5 kg 0.5 kg
Q 100° C F
Q 20° C t° C
Q =Q
(mc∆T) = (mc∆T)
F F 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 F
F 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 F www.schoolDD.com 7

9. 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
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 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
20° C Q
Q
2 kg 2 kg
20° C Q t° C
Q =Q
(mc∆T) = (mc∆T) + (mc∆T)
F F 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 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 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
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 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 F www.schoolDD.com 9

11. t = -12.39° C F F F 0° C
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 0° C 0.15 kg Ans
10.1.7 F F
3
1. F
- F F F F F F
F F F
2. F
- F F 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 F www.schoolDD.com 10

12. 10.2 F
- F F F F F F
F F
10.2.1 F
F F F
P
T
V PV = F
10.2.2 F
F F F
V
P
V T
= F T
10.2.3 F
F F F
V
= F
F F F
F = 1 2 (T F )
F F P, V F F = F F F F T
F K F
– F F www.schoolDD.com 11

13. F
F n n= F =
1
= F F F 1 2
F
F (T F )
F
PV = nRT R= F F = 8.31 J/mol K
. F
n=
= 6.02 x 1023
F 1 F 6.02x1023
F PV = RT
F F KB = = F F F = = 1.38x10-23 J/K
F PV = NKB T
PV = nRT PV = NKB T F F F F
F F F F
1. F PV = nRT
2. F F F P T, P n, V T, V n P F
3. F F F F F 1 2 F F F
F F
= R = R
= (P F P F )
F F
– F F www.schoolDD.com 12

14. F 10
F 1 F 2 x 105 Pa 7˚ C F F F F
2 F F 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
=
P n
F =
=
T2 = 560 K 560 273 = 287° C Ans
F 11
F F 105 Pa F 27˚ C -3˚ C
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
=
V n
– F F www.schoolDD.com 13

15. F =
=
P2 = 0.9x105 Pa Ans
F 12
F F F 17˚ C F 2 x 105 Pa F
F 27˚ C F 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
=
V n
F =
= F P FF P F F
=
Pg 2 = 2.09x105 Pa Ans
F 13
F 10 27˚ C 105 Pa F
F
– F 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
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 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
Px20x10-3 = x8.31x300 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
=
F n= M
F =
=
P2 = 1.10 atm Ans
– F F www.schoolDD.com 16

18. F 17
F F F F
F
V1 V2 = V1
P1 P2 = ½ P 1
T1 T2 = T1
n1 n2 = ?
F PV = nRT
F F 2
=
F V T
F =
=
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
3
1 = 10 kg/m 2=?
F PV = nRT
F F 2
=
F n= M
– F F www.schoolDD.com 17

19. F =
=
=
=
= 5.17 kg/m3 Ans
10.3 F F
10.3.1 F
F
1. F F F
F F F
2. F F F
3. F F F F
10.3.2 F ( < EK > )
F F < EK > F F
1 F (J)
< EK > = KB T
F F N F N< EK >
N< EK > = NKB T
= PV ( PV = NKB T )
= nRT ( PV = nRT )
- < EK > F F F F F
– F F www.schoolDD.com 18

20. 10.3.3 F
F F F vrms (vroot mean square )
F F (m/s)
vrms = M= F 1 (kg)
- 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 > = ?
< EK > = KB T KB = = = 1.38x10-23
= x 1.38x10-23x300
< EK > = 6.21x10-25 J Ans
. N< EK > = ?
. N N = n NA
= = x 6.02x1023
N = 6.02x1024
N< EK > = 6.02x1024x 6.21x10-25 = 3.73 J Ans
– F F www.schoolDD.com 19

21. . vrms = ?
vrms =
= =
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
F F N< EK > = NKBT = PV ( PV = NKBT)
-3
= x2x10x10
N< EK > = 3x10-2 J Ans
F < EK > < EK > = =
< 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> = ?
F < EK > < EK > = KBT
F N< EK > = NKBT = PV ( PV = NKBT)
< EK > =
= x
< EK > = 7.5x10-21 J Ans
– F F www.schoolDD.com 20

22. F 22
F
. F F 2 F F F
. F F 2 F F F
.
T1 T2 = 2 T1
<EK1> <EK2> = ?
< EK > = KBT
F < EK > T
= =
< EK2 > = 2< EK1 > Ans
.
V1 V2 = V1
P1 P2 = 2 P 1
N1 N2 = N1
<EK1> <EK2> = ?
< EK > = KBT
N< EK > = NKBT = PV
F < EK > P N V
= =
< EK2 > = 2< EK1 > Ans
F 23
F A 1 7˚ C F B 2 27˚ C
F F 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 F www.schoolDD.com 21

23. N< EK > N< EK > = NKBT
= nRT
(N< EK >)A + (N< EK >)B = ( nRT)A + ( nRT)B
= R(nATA + nBTB)
= x8.31(1x280 + 2x300)
F = 1.097x104 J Ans
F 24
F F F ( MH2 = 2 g/mol, MO2 =
32 g/mol)
H2
vrmsH2 O2
vrmsO2
vrms =
F vrms T
= = = 4
vrmsH2 = 4 vrmsO2 Ans
F 25
F F F 4 F
vrms F
=4
vrms1 vrms2
vrms =
PV = nRT
F RT =
vrms =
– F F www.schoolDD.com 22

24. = = =
F vrms P
= = = 2
vrms2 = 2 vrms1 Ans
10.4
-
- F FF F F
F U F F
F F F F F
F F F F F F F
F F
U = N< EK >
F FF U = NKBT
= nRT
= PV
10.4.1 F 1 F F
F F F F
〉 ∆W
∆Q = ∆U + ∆W
〉 ∆U
〉 ∆Q
ΔW = FΔs
= PAΔs
= PΔV
– F 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 + ) 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
= NKB∆T + P∆V
= x1024x1.38x10-23x1 + 0
∆Q = 20.7
F F F F F ∆Q = 20.7 J Ans (∆Q +)
– F 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
= nR∆T + (-F∆s) ( F )
= x1x8.31x1 – 100x0.20
∆Q = -7.5 ( F )
F ∆Q = 7.5 J Ans
F 29
F F F F F F F
P
∆W
∆U ∆V
〉 ∆Q = ?
∆Q = ∆U + ∆W
= NKB∆T + P∆V
= P∆V + P∆V ( PV = NKBT)
∆Q = P∆V
F F F F F ∆Q = P∆V J Ans
– F F www.schoolDD.com 25

27. F 30
F F F 20 27˚ C 105 Pa F
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 = ?
V2 F =
P n
F =
=
V2 = 22
∆Q = ∆U + ∆W
= NKB∆T + P∆V
= P∆V + P∆V ( PV = NKBT)
= P∆V = x10 x(22-2)x10-3
5
∆Q = 500
F F F F F ∆Q = J Ans
– F F www.schoolDD.com 26