Hwa Chong Institution
Sec 3 Physics (SMTP)
Chapter 4: Thermal Properties
Assignment 5
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Answer
1(a)
2 kg of pure substance X was heated uniformly from its solid state from a
temperature of 25.0 ºC using a 100 J/s heater to obtain its heating curve.
The temperature of X was taken in intervals of 5.0 minutes until it reaches the gaseous
state and tabulated as shown in the figure below. Assume that the heat energy supplied
was constant and no heat energy was lost during the heating process.
Melting point of pure X = 40.0 ºC
Boiling point of pure X = 70.0 ºC
Time when X began melting = 2.5 minutes
Time taken for all of solid X to melt = 5.0 minutes
Time when X began boiling = 10.0 minutes
Specific latent heat of vaporization of X = 30.0 kJ/kg
Time
/ min
0.0
5.0
10.0
15.0
20.0
25.0
Temperature
/ ºC
25.0
40.0
70.0
70.0
70.0
85.0
1m
1m
1m
25
(i)
By analyzing the data obtained and using the given information, plot the heating
curve of pure substance X in the grid lines provided.
[3]
1
(ii)
Calculate the specific heat capacity of the solid X.
Heat energy supplied, Q
Temperature change, ΔT
[2]
= 100 x (2.5 x 60)
= 15 kJ
[1]
= 40 – 25
= 15 ºC
[1]
Specific heat capacity of solid Z, csolid
= Q/ mΔT
= 15 000 / (2 x 15)
= 500 J/kgºC
2
[1]
(b)
Figure below shows the cooling curve graphs of two pure liquids, Y and Z, of the
same mass.
Temperature
T1
Y
Fig. 11.2
Z
Time
0
(i)
Which substance has a greater specific heat capacity in the liquid state?
Explain your answer clearly.
Substance Y.
[3]
[1]
For the same cooling condition, fall in temperature for substance Y is
less than substance Z.
[1]
This indicates that a greater amount of heat energy needs to be lost by
substance Y compared to Z for the same amount of fall in temperature.
[1]
(ii)
Which substance has a greater specific latent heat of fusion?
Explain your answer clearly.
[3]
Substance Z
For the same mass, substance Z takes a longer time to change state
indicating that greater amount of latent heat needs to be lost by Z compared
to Y to change from liquid to solid state
3
2
Figure below shows an experimental set-up used for determination of the specific heat
capacity of water and the specific latent heat of steam.
The inner copper can has a mass of 300 g when empty. It contains 240 g of water. The
copper can has a specific heat capacity of 0.4 J oC-1 g-1 and its outer surface is polished.
The air between the copper can and the metal container is trapped by a layer of
styrofoam. The circuit of the heater is switched on continuously while the readings of
the ammeter, the voltmeter, the thermometer and the chemical balance are recorded at
regular time intervals. The graphs of the total mass against time and temperature
against time have been plotted as shown below:
Boiling starts
4
(a)
Assume the specific heat capacity of water is c, write an expression for
the heat energy gained by the water from time 10 min to 20 min, mass of the
water being 240 g. Find also the heat energy gained by the copper can in the
same time.
[3]
Let specific heat capacity of water be cwater Jg-1oC-1
Heat energy supplied, Q = mwater . cwater . ΔT
Q = 240 . cwater . . (40 - 20)
= 4800. cwater
[1]
Heat energy supplied, Q = mcopper . ccopper . ΔT
Q = 300 . 0.4 . (40 - 20)
= 2400 J
[1]
[1]
(b)
The heater produced heat energy at a rate of 42 J/s constantly. Calculate the
heat energy given out by the heater from time = 10 min to 20 min.
[2]
Qsupplied
= 42 x 10 x 60
[1]
= 25200 J
[1]
(c)
Calculate the specific heat capacity of water using the results of (a) and
(b).
[2]
ΔQ=0
Q water + Q copper + Q heater = 0
240 . cwater . . (40 - 20) + 2400 + (-25200) = 0
-1
-1
cwater
= 4.75 Jg ºC
[1]
[1]
Note:
Energy from the heater is denoted as negative as the heater “loses” heat
energy to the system.
5
(b)
Explain, using the kinetic theory of matter, why the temperature of water
does not change from time = 70 min onwards.
[1]
The heat energy supplied is used to weaken the intermolecular bonds in
water (causing a change of state), not to raise the temperature of water.
Note:
Do not accept “break” intermolecular bonds
(c)
Calculate the specific latent heat of steam.
[2]
Heat energy supplied, Q = mlv
42 x (90 – 70) x 60
= ( 980 – 958) x lv
42 x 20 x 60 = 22 x lv
lv = 2.3 x 103 J g-1
(d)
[1]
[1]
State the advantage of trapping air outside the copper can.
To reduce loss of heat energy from the copper can to the surroundings
since air is a poor conductor of thermal energy.
[1]
6
[1]
3.
Read the following passage about a steam wax extractor and answer the questions that
follow.
Natural bee honey is a valuable gift from nature. It is taken for its natural anti-aging
properties.
Honey is a component of raw beeswax which is taken from the beehive.
Beeswax melts at 63C and its specific latent heat of fusion is 1.7  105 J/kg which is only
about 8% specific latent heat of vaporization of water. A steam wax extractor is a simple and an
effective tool to extract wax from a beehive, as beeswax absorbs vapour and becomes molten wax
which is collected by a collector pan.
A steam wax extractor consists of a water chamber in order to boil the water which is heated
by gas or electricity. The collector pan allows the molten wax and condensed steam to run out into
a collecting vessel. There is a skirt above the spout collector that deflects any molten wax into the
collector rather than falling into the boiling water below.
7
(a)
Suggest two reasons why steam is a better medium to melt beeswax than boiling water.
[2]
Steam is a gas which can diffuse into the wax fragments easily / heat up the wax
fragments uniformly.
[1]
Steam releases more heat energy to the wax fragments than water / less water is
needed to melt wax using the steam method.
[1]
(b)
Define the term specific latent heat of fusion.
[1]
Specific Latent heat of fusion is the quantity of heat energy absorbed to melt 1kg of
a substance without a change in temperature.
OR
Specific Latent heat of fusion is the quantity of heat energy released to freeze 1kg
of a substance without a change in temperature.
(c)
(d)
Using the information above and given that the specific latent heat of vaporization and
specific heat capacity of water are 2.26  106 J/kg and 4200 J/kg respectively, estimate
the maximum mass of beeswax melted by 0.1kg of steam at 100C.
[4]
Heat energy released by steam
=
=
=
Mass of beeswax melted
Q / lf
241 540 / 1.70 x 105
1.4 kg
[1]
=
=
=
mlv + mc
(0.1 x 2.26 x 106) + (0.1 x 4200 x (100 – 63)) [1]
241 540 J
[1]
Describe what happens to the wax molecules during melting.
8
[1]
[2]
Wax molecules at the gain energy from steam molecules.
[1]
They vibrate about their equilibrium positions so vigorously that they break the
intermolecular bonds and become mobile and free to move around, forming a
liquid.
[1]
(e)
Suggest one way to increase the efficiency of extracting wax using this extractor. [1]
Cover the top of the extractor to prevent any steam from escaping
OR
Use lagging/insulating material to wrap the extractor to reduce loss of heat energy to
the surroundings.
4
Read the following passage about icebergs and answer the question that follow:
Icebergs are huge floating pieces of ice formed along the coast of
Greenland. About 40000 medium to large icebergs are formed
annually. They are taken by ocean currents as far south as 48o
north latitude.
The typical mass of an iceberg is about 1 x 104 tons to 1 x 107
tons and the biggest iceberg ever recorded had a mass of 1.4 x
1012 tons. The interior temperature of an iceberg is about -15oC
to 20oC while the surface temperature rises to 0oC.
Some scientists suggested towing some icebergs to the Middle
East to solve the problem: water shortage. However, an iceberg
is too massive to tow. In addition, it would take too long to reach
the Middle East because a large iceberg may only last for about
11 days in 10oC water.
Given: 1 ton = 1000 kg
Specific heat capacity of ice = 2100 J/kg.oC
Specific heat capacity of water = 4200 J/kg.oC
Specific latent heat of fusion of water = 3.34 x 105 J/kg
(a)
Explain why the surface temperature of an iceberg is always 0oC.
The surface of the iceberg is melting and the melting point of ice is 0oC.
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[1]
(b)
Describe how the molecules in an iceberg gain energy from warm seawater.
[2]
Molecules at the surface of the iceberg are collided into by water molecules which
are moving at higher average velocity.
[1]
By collision, some kinetic energy of the water molecules is transferred to the
surface molecules on the iceberg.
[1]
(c)
Assume 75% of an iceberg is at -20oC and the remaining 25% is at 0oC. Estimate the
total thermal energy required to completely melt an iceberg of mass 1 x 107 tons.
[3]
Q
= m1c∆θ +m2  f
= (0.75 x 1x107 x 1000)(2100)(20-0) + (1x107 x 1000) (3.34 x 105)
[2]
= 3.66 x 1015 J
(d)
[1]
Suggest one method to extend the time an iceberg could be kept in warm seawater.
Explain how this method can help to extend the time.
[2]
Cover the upper part of an iceberg with a silvery sheet. [1]
This reduces loss of thermal energy by radiation. [1]
The End
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