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Physics
WAEC / WASSCE 2013
Thermal Physics
Which of the following types of thermometer is used for the calibration of other thermometers?
A
liqud-in-glass thermometer
B
constant volume gas thermometer
C
optical pyrometer
D
thermocouple
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #35665)
Physics
WAEC / WASSCE 2013
Thermal Physics
The magnitude of the expansion or contraction of a substance depends on the i. temperature change ii. nature of the substance iii. size of the substance. Which of the statements above are correct?
A
i and ii only
B
ii and iii only
C
i and iii only
D
i, ii and iii
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #35666)
Physics
WAEC / WASSCE 2013
Thermal Physics
Solution
Ice of mass 10g at 5^\circC was completely converted to water at 0^\circC, calculate the quantity of heat used. [specific heat capacity of ice = 2.1 Jg\(^{-1}\)K\(^{-1}\), specific latent heat of fusion of ice = 336 Jg\(^{-1}\)]
A
105J
B
3255J
C
3465J
D
16821J
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Q = ML + MC\(\theta\)
= 10 x 335 + 10 x 2.1 x 5
= 3465J
Source: Myschool Past Questions Archive (Record #35669)
Physics
WAEC / WASSCE 2013
Thermal Physics
Solution
When salt is dissolved in water, the freezing point of the water
A
increases
B
decreases
C
remains the same
D
increases and later decrease
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The salt in the water (same for any other impurities) lowers the freezing point of the solution so that water freezes at a lower temperature.
Source: Myschool Past Questions Archive (Record #35671)
Physics
WAEC / WASSCE 2013
Thermal Physics
Solution
Which of the following radiations has it's frequency lower than that of infrared radiation?
A
Utra-violet rays
B
gamma rays
C
x-rays
D
radio waves
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Radio waves have lower frequencies compared to infrared radiation, while ultraviolet rays, gamma rays, and x-rays have higher frequencies.
Source: Myschool Past Questions Archive (Record #35680)
Physics
UTME / JAMB 2012
Thermal Physics
Clinical thermometers are examples of
A
pressure gas thermometer
B
resistance thermometer
C
alcohol thermometer
D
mercury- in - glass thermometer
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #27390)
Physics
UTME / JAMB 2012
Thermal Physics
Solution
Two metals P and Q are heated through the same temperature difference. If the ratio of the linear expansivities of P to Q is 2:3 and the ratio of their
length is 3: 4 respectively, the ratio of the increase in lengths of P to Q is
A
1:2
B
2:1
C
8:9
D
9:8
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
P: \(\frac{\alpha h \theta}{\alpha h \theta}\)
\(\frac{2 \times 3 \times \theta}{3 \times 4 \times \theta}\)
\(\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #27392)
Physics
UTME / JAMB 2012
Thermal Physics
Solution
Calculate the temperature change when 500J of heat is supplied to 100g of water
[specific heat capacity of water = 4200JKg\(^{-1}\)k\(^{-1}\)]
A
12.1oC
B
2.1oC
C
1.2oC
D
0.1oC
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(\theta = \frac{Q}{MC}\)
\(\theta = \frac{500}{0.1 \times 4200}\) ≈ 1.2 ºC
Source: Myschool Past Questions Archive (Record #27395)
Physics
WAEC / WASSCE 2012
Thermal Physics
Which of the following substances is the most volatile at room temperature?
A
water
B
diesel
C
petrol
D
kerosine
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #35528)
Physics
WAEC / WASSCE 2012
Thermal Physics
Diagram
The diagram above illustrates trapped air in a syringe placed in water that is gradually heated. It is observed that the piston rises as the temperature of the water rises. Which of the following statements explains this observation?
Attached Figure / Diagram
A
frequency of collision of trapped air molecules with walls of the syringe is more
B
molecules of trapped air expands more
C
speed of the trapped air molecules decreases
D
number of molecules of trapped air increases
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #35518)
Physics
WAEC / WASSCE 2012
Thermal Physics
Solution
A metal sheet of area 100cm\(^2\) was heated through 70^\circC. Calculate its new area if the linear expansivity of the metal is 0.000017K\(^{-1}\).
A
100.06 cm2
B
100.12cm2
C
100.24cm2
D
100.36cm2
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Formula: A\(_2\) = A\(_1\)(1 + \(\beta \theta\))
where \(\beta\) = The area expansivity
\(\theta\) = Change in temperature.
\(\beta\) = 2\(\alpha\)
where \(\alpha\) = linear expansivity of the body.
⇒ \(\beta\) = 2 x 0.000017
= 0.000034 K\(^{-1}\)
\(\therefore\) A\(_2\) = 100 (1 + (0.000034 x 70))
= 100(1 + 0.00238)
= 100(1.00238)
= 100.238 cm\(^2\)
= 100.24 cm\(^2\) (to 2 d.p)
Source: Myschool Past Questions Archive (Record #35565)
Physics
WAEC / WASSCE 2012
Thermal Physics
Solution
A piece of copper of mass 20g at a temperature of 110^\circC was dropped into a mixture of ice and water at 0^\circC. If the final steady temperature of the mixture is 0^\circC . Calculate the amount of ice that melted [Specific heat capacity of copper = 0.4 Jg\(^{-1}\)K\(^{-1}\), specific latent heat of fusion of ice = 330Jg\(^{-1}\)]
A
0.37g
B
0.60g
C
2.40g
D
2.70g
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Mice = \(\frac{MC}{\theta}\)
= \(\frac{20 \times 0.4 \times 110}{330}
Heat lost by copper = Heat gained by ice
m\(_{copper}\) c \(\theta\) = m\(_{ice}\) L
20 x 0.4 x (110 - 0) = m\(_{ice}\) x 330
m\(_{ice}\) = \(\frac{20 \times 0.4 \times 110}{330}\)
= \(\frac{8}{3}\)
= 2.67 g
= 2.70g (to 1 d.p)
Source: Myschool Past Questions Archive (Record #35581)
Physics
UTME / JAMB 2011
Thermal Physics
Solution
I. High thermal capacity.
II. High sensitivity.
III.Easy readability.
IV. Accuracy over a wide range of temperature
From the statement above, the qualities of a good thermometer are
A
I and II
B
I, II,III and IV
C
I, III and IV
D
II, III and IV
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
A good thermometric liquid should not have a high thermal capacity so that the themperetue of the body being measured will not be affected
∴ II, III and IV are qualities of a good thermometer
Source: Myschool Past Questions Archive (Record #14380)
Physics
UTME / JAMB 2011
Thermal Physics
Solution
The ground is always cold at night because the
A
atmosphere absorbs the sun's energy at night
B
earth radiates heat to the atmosphere at night
C
sun no longer shine at night
D
atmosphere reflects the sun's energy at night
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The earth radiate heat to the atmosphere at night thereby lowering the temperature of the ground
Source: Myschool Past Questions Archive (Record #14404)
Physics
UTME / JAMB 2011
Thermal Physics
Solution
A block of aluminum is heated electrically by a 30 w heater if the temperature rises by 10°c in 5 minutes, the heat capacity of the aluminum is
A
900 JK-1
B
90 JK-1
C
100 JK-1
D
200 JK-1
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Heat supplied = Heat gained Pt = MCDθ
30 x 5 x 60 = c x 10
where c = mc which is the heat capacity
c = (30 x 5 x60)/10 = 900 JKg-1
Source: Myschool Past Questions Archive (Record #14420)