Download Physical Quantities, Kinetic Theory, Effects of

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Thermal power station wikipedia , lookup

Heat pump wikipedia , lookup

Energy policy of the European Union wikipedia , lookup

Solar thermal energy wikipedia , lookup

Energy storage wikipedia , lookup

Geothermal heat pump wikipedia , lookup

Alternative energy wikipedia , lookup

Energy Independence and Security Act of 2007 wikipedia , lookup

Internal energy wikipedia , lookup

Solar water heating wikipedia , lookup

Conservation of energy wikipedia , lookup

Cogeneration wikipedia , lookup

Compressed air energy storage wikipedia , lookup

Energy applications of nanotechnology wikipedia , lookup

Environmental impact of electricity generation wikipedia , lookup

Economizer wikipedia , lookup

Transcript
38
Thermal Physics Term Test
Name : _______________________(
)
Class : _________
Date : ______
Physical Quantities, Kinetic Theory, Effects of Thermal Energy,
Thermal Properties of Matter
Section A: 5 marks
1
The diagram shows part of a micrometer screw gauge.
What is the reading shown?
A 5.31 mm
C 5.81 mm
2
B 5.79 mm
D 6.31 mm
A copper plate is heated to 100oC and then allowed to cool. It cools by emitting
A electrons
C infra-red radiation
3
B gamma radiation
D ultraviolet radiation
Hot water at 100°C is added to 5 g of ice at 0°C. What is the minimum mass of
hot water needed to melt the ice?
[take the s.l.h. of fusion of ice to be 336 J/g and s.h.c. of water to be 4.2 J/(g K)]
A 4.0 g
4
B 16.8 g
C 70.6 g
D 400 g
A substance is heated at a constant rate and a graph of temperature against time is
plotted. Which parts of the graph correspond to a mixture of liquid and gas state?
A
B
C
D
E
PQ
QR
RS
ST
TU
1 of 6
5
The diagrams represent two blocks of copper, P and Q, each receiving the same
amount of energy W. The mass of P is twice the mass of Q. The temperature rise
of P is half the temperature rise of Q.
Which statement about P and Q is correct?
A
B
C
D
The heat capacity of P is half the heat capacity of Q
The heat capacity of P is twice the heat capacity of Q
The s.h.c. of P is half the s.h.c. of Q
The s.h.c. of P is twice the s.h.c. of Q
Section B: 33 marks
6.
Simplify the followings to the appropriate unit and significant figure:
(i)
3.9 kg x 24.0 m 4
0.015 m 2
= 6.2 x 103 kgm2 (2 sf)
1
(ii)
[2]
(5.28 m/s) 2
7.20 m/s 2
= 3.87 m (3 sf)
When you step out of a warm shower, you feel cold as the water on your body
begin to evaporate. Explain this effect using kinetic theory.
[3]
Water molecules on the skin gained thermal energy from my body.
The thermal energy gained by the molecules is converted to kinetic energy.
[1]
Some water molecules gained enough kinetic energy to escape the water
surface as vapour. Hence, as evaporation occurs, my body feels cool. [1]
As the more energetic molecules leave the liquid, the average kinectic energy
of the molecules in the liquid decreases. This cause the temperature of the
liquid to drop.
2 of 6
2
A student said that a piece of paper would fall on ground when released because
the atmospheric pressure forced the paper to fall down.
Using kinetic theory, determine whether the above statement is true.
[3]
Atmospheric pressure is caused by the average force per unit area exerted by
the air molecules as they bombard the paper.
Since there are air molecules colliding with both the upper and lower side of
the paper, there will be no net force exerted on the paper by the colliding air
molecules.
Hence, her statement is not true.
3
A pack which contains a frozen liquid is placed in a holder under the lid of a box
as shown. The box is then used to keep food cold.
(a)
Suggest why the frozen pack is placed at the top of the box rather than at
the bottom.
[2]
Air around the pack gets cooled and sinks to the bottom of the box.
Warmer air at the bottom hence rises up.
[1]
Hence, convection current is set up to continuously cool the air in the
box.
[1]
(b)
Explain why food can be kept cold longer if the walls are made of rigid
plastic rather than of metal;
[2]
Plastic is a poor conductor of thermal energy.
while metal is a good conductor of thermal energy. [1]
If the wall is made of plastic, transfer of thermal energy will be
through the walls from the surroundings into the
box will be slower.
3 of 6
(c)
What colour should the box be so as to be most effective in keeping food
cold? Explain your answer.
[2]
The box should be painted silver.
[1]
Silver surfaces are poor emitters and absorbers of radiant energy.
[1]
(d)
Suggest one more improvement in the design of the box so that it will be
more effective in keeping food cold. Explain how the improvement works.
[3]
The walls of the box can be made double layer, with air in between.
[1]
Air is a bad conductor of heat energy.
[1]
Hence, less thermal energy will be conducted through the walls
from the surroundings into the box.
[1]
4
In the experiment shown, the ice remains intact for several minutes as heating
progresses. Explain how this can be so.
[2]
Heat energy is applied at the top of the test tube. Hence, no convection
current is set up to transfer thermal energy from the heat source to ice.
Also, water is a poor conductor of thermal energy. Hence, thermal energy is
not effectively transferred from the heat source to ice.
4 of 6
5
Explain why burns caused by steam on the skin are often more severe than burns
caused by boiling water, even though both are at 100oC.
[2]
For the same mass, steam contains more heat energy than boiling water.
[1]
The extra energy in steam comes from its latent energy of vaporisation.
[1]
6
The diagram below shows a simple shell and tube heat exchanger.
Water is passed through the shell at the rate of 10kg/min to cools fluid Y in
the tubes. The water temperature rises from 40oC to 85oC
(take specific heat capacity of water as 4200 Jkg-1oC-1)
(a)
Find the heat energy lost from fluid Y in 1 second
Heat energy lost by fluid Y
= heat energy gained by water
= mc (Tfinal – T initial)
= 10/60 (4200) (85 – 40)
= 31500 J
= 3.2 x 104 J (2sf)
(b)
[2]
[1]
[1]
State an assumption you made for part (a).
All the heat energy lost by fluid Y is gained by water.
[1]
5 of 6
(c)
Give two reasons why water is often used as a cooling agent.
[2]
The two reasons are:
1) Water has a high specific heat capacity which means it can gained
a large amount of heat energy for a small increase in temperature.
2) Water is cheap and readily available.
7
Figure 8 (a) and (b) shows the experimental set-ups used to measure the heat
capacity of a liquid and the heat capacity of a metal respectively.
Figure 8(a)
Figure 8(b)
Figure 8(c) shows the temperature-time graph obtained in an experiment to
determine the specific heat capacity of aluminium by using the apparatus in
figure 8 (a) and (b).
Figure 8(c)
6 of 6
The mass of the aluminium is 0.50 kg and the heater is switched on for three
minutes.
(a)
What is the purpose of the stirrer shown (in figure 8a)?
[1]
It is to stir the liquid so that the heat energy supplied by the heater
will be absorbed by all parts of the liquid uniformly.
(b)
(c)
Is the apparatus in Figure 8a suitable to find out the specific heat capacity
of a flammable liquid? Explain.
[2]
No.
[1]
The heater is in direct contact with the liquid.
Hence, a flammable liquid may catch fire.
[1]
What is the purpose of the oil in Figure 8b?
[1]
The oil improves thermal contact between the heater and the metal
block.
(d)
Account for the slow rate of temperature rise initially in the graph [1]
The metal block has not been evenly heated.
(e)
If the heat energy supplied by the heater in 1 second is kept constant at
24J, calculate the specific heat capacity of aluminium.
[2]
Heat supplied by heater
= mcaluminium (Tfinal – T initial)
24 (3) (60)
= 0.50 caluminium (35 -25)
[1]
caluminium
= 864 J kg-1oC-1
[1]
7 of 6
Bonus Question:
8
At a reception, a 0.5 kg chunk of ice at -10oC is placed in 3.0 kg of “iced” tea at
20oC.
In what state will the final mixture be? Show your calculations.
[4]
(cwater = 4186 Jkg-1oC-1, cice = 2100 Jkg-1oC-1, Lf of water = 333 kJkg-1)
Note: Tea can be considered as water.
Heat energy released by tea (20oC to 0oC) = 3(4186) (0 – 20)
= - 251160
= 250 kJ
Heat required by ice (-10oC to 0oC)
Heat required by ice (melting)
[1]
= mcice 
= 0.5 (2100) (0 – (-10))
= 10500 J
= 10.5 kJ
[1]
= mLf
= 0.5 (333) 1000
= 166.5 kJ
[1]
Total heat energy required to melt ice = 10.5 + 166.5
= 177 kJ
177 kJ is not enough to bring 3.0 kg of water at 20oC down to 0oC.
Hence, the mixture must end up all water, somewhere between 0oC and 20oC.
[1]
- End of Test -
8 of 6