PHYSICS LAB %
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Dec 6, 2023
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ENG TECH 1PH3
MCMASTER – MOHAWK JOINT VENTURE
BACHELOR OF TECHNOLOGY PARTNERSHIP
FOUR-
YEAR UNIVERSITY DEGREE PROGRAMS
EXPERIMENT NO:
5
TITLE:
LATENT HEAT
Submitted by:
Sofiya Sniezko
Lab Section:
L01
Partner:
Instructor:
Dr. Reza Ushaksaraei
Date lab performed:
November 9, 2023
Date of submission:
November 23, 2023
ENG TECH 1PH3
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1
PURPOSE:
To study the latent heat of fusion and the latent heat of vaporization for water.
RESULTS
TABLE
5.1
WEIGHT OF
EMPTY
STYROFOAM
CUP
(g)
TEMP OF
COOLED
BLOCK
(°C)
WEIGHT
OF
METALLIC
BLOCK (g)
TEMP. OF
HEATED
WATER
(°C)
WEIGHT OF
STYROFOAM
CUP
WITH CONTENTS
(g)
FINAL
TEMP
OF
MIXTURE
IN CUP
(°C)
1.77
0
70.70
54.0
106.74
46.00
TABLE
5.2
WEIGHT OF
EMPTY
STYROFOAM
CUP
(g)
TEMP OF
ICE
(°C)
WEIGHT OF
STYROFOAM
CUP WITH ICE
(g)
TEMP. OF
HEATED
WATER
(°C)
WEIGHT OF
STYROFOAM
CUP
WITH MIXTURE
OF
WATER AND ICE
(g)
FINAL
TEMP
OF
MIXTURE
IN CUP
(°C)
1.77
0
27.18
56.0
114.16
27.80
TABLE
5.3
WEIGHT OF EMPTY
STYROFOAM
CUP
(g)
TEMP OF
COOLED WATER
(°C)
WEIGHT OF
STYROFOAM
CUP
WITH WATER (g)
WEIGHT OF
STYROFOAM CUP
WITH MIXTURE OF
WATER + STEAM (g)
FINAL TEMP
OF
MIXTURE
IN CUP (°C)
1.77
1.5
105.58
110.20
29.5
CALCULATIONS:
1)
T
b
= 0 °C
T
f
= 46.00 °C
T
w
= 54.00
°C
m
b
= 70.70 g
= 0.07070 kg
m
w
= 106.74g -70.70g -1.77g
ENG TECH 1PH3
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2
= 34.27g
= 0.03427 kg
c
w
= 4186 J/(kg
°C)
c
b
=
-
m
w
c
w
(T
f
-T
w
)
m
b
(T
f
-T
b
)
= -
0.03427kg x
4186 J/(kg
°C) x (46.00
°C – 54.00°C)
0.07070kg x (46.00°C - 0°C)
= 352.9
J/(kg
°C)
2) m
w
= 114.16g – 27.18g
= 86.98 g
= 0.08698 kg
m
i
= 27.18 – 1.77g
= 25.41 g
= 0.02541 kg
c
w
= 4186 J/(kg
°C)
ΔT
w
=
L
f
= - m
w
c
w
ΔT
w
- m
i
c
w
ΔT
i
m
i
Latent Heat of Vaporization of Water
The steam condenses and releases its latent heat of vaporization into the cold water and the
condensed steam cools adding further heat to the cool water in the calorimeter causing and
increase in temperature.
The heat that is added to cause vaporization represents an increase in heat, Δ
Q
v
= m
v
L
v
,
which
is a positive value. When the vapour condenses, the change in heat of the vapour is negative
since the vapour has given up its latent heat, Δ
Q
v
= -m
v
L
v
.
The heat lost by the steam is equal
to the heat gained by the water Δ
Q
w
= -
Δ
Q
v
.
The relationship needed to determine these heats
is based on changes (Δ) in heat. The heat gained by the water is
m
w
c
w
Δ
T
w
.
The heat lost by the
vapour as it condenses and cools down to the final temperature is -
m
v
L
v
+ m
v
c
w
Δ
T
v
,
where
Δ
T
v
= (T
f
-
100) .
Note:
The temperature at which steam condenses is assumed to be 100°C.
Since Δ
Q
w
= -
Δ
Q
v
,
it follows that
m
w
c
w
Δ
T
w
= - (m
v
L
v
+ m
v
c
w
Δ
T
v
)
and
ENG TECH 1PH3
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3
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