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Physics
WAEC / WASSCE 1996
Thermal Physics
Solution
What is the difference in the amount of heat given out by 4kg of steam and 4kg of water when both are cooled from 100oC to 80oC? (Specific latent heat of steam = 2.260,000J kg-1) (Specific heat capacity of water =4,200 J kg-1)
A
336,000J
B
672,000J
C
8,704,000J
D
9,040,000J
E
9,376,000J
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(Q_{s} - Q_{w} = (4 \times 2260000) - (4 \times 4200 \times (100 - 80))\)
= \(9040000 - 336000\)
= \(8,704,000 J\)
Source: Myschool Past Questions Archive (Record #25215)
Physics
WAEC / WASSCE 1996
Thermal Physics
Water boils at a lower temperature when heated at the top of a mountain than at sea-level because at the top of the mountain the
A
saturation vapour pressure of water is higher than at sea-level
B
sun adds more heat to the water
C
temperature is lower than at sea- level
D
relative humidity of the atmosphere is lower
E
pressure of the atmosphere is lower
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #25220)
Physics
WAEC / WASSCE 1996
Thermal Physics
Which of the following substances is most viscous at room temperature?
A
Water
B
Alcohol
C
Petrol
D
Palm oil
E
Kerosine
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 #25253)
Physics
WAEC / WASSCE 1995
Thermal Physics
A mercury-in-glass thermometer reads - 20o at the ice point and 100o at the steam point. Calculate the celcius temperature corresponding to 70o on the thermometer.
A
41.0oC
B
50.0oC
C
62.5oC
D
70.0oC
E
75.0oC.
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #25063)
Physics
WAEC / WASSCE 1995
Thermal Physics
A brass rod is 2 m long at a certain temperature. Calculate the linear expansion of the rod for a temperature change of 100K. (Take the linear expansivity of brass as 1.8 x 10'5K").
A
0.3600 m
B
0.1800 m
C
0.0360 m
D
0.0036m
E
0.0018 m
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 #25064)
Physics
WAEC / WASSCE 1995
Thermal Physics
Solution
Hot water at a temperature of t is added to twice that amount of water at a temperature of 30^\circC. If the resulting temperature of the mixture is 50^\circC. Calculate t.
A
90oC
B
80oC
C
50oC
D
40oC
E
30oC.
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Final Temperature = \(\frac{m_1 T_1 + m_2 T_2}{m_1 + m_2}\)
where m\(_1\) and m\(_2\) are the weights of the water in the first and second containers.
T\(_1\) and T\(_2\) are the temperatures of the water in the first and second container.
Given Final temperature = 50^\circC;
Let m\(_1\) = b, then m\(_2\) = 2b.
T\(_1\) = t; T\(_2\) = 30^\circC
\(\therefore 50 = \frac{b(t) + 2b(30)}{b + 2b}\)
\(50 = \frac{bt + 2b(30)}{3b}\)
\(50(3b) = bt + 2b(30) \implies 150b = b(t + 60)\)
\(150 = t + 60 \implies t = 150 - 60 = 90^\circC\)
Source: Myschool Past Questions Archive (Record #25066)
Physics
WAEC / WASSCE 1995
Thermal Physics
Solution
The temperature of glass vessel containing 100cm3 of mercury is raised from 10oC to 100oC. Calculate the apparent cubic expansion of the mercury. (Real cubic expansivity of mercury = 1.82 x 10-4K-1) (Cubic expansivity of glass = 2.4 x 10-5K-1)
A
0.52cm3
B
1.42cm3
C
1.87cm3
D
5.22cm3
E
14.22cm3
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
To calculate the apparent cubic expansion of the mercury, we use the formula for apparent expansion:
\(\text{Apparent Expansion} = V_0 \cdot \left( \beta_m - \beta_g \right) \cdot \Delta T\)
Where:
- \( V_0 = 100 \, \text{cm}^3 \) (initial volume of mercury)
- \( \beta_m = 1.82 \times 10^{-4} \, \text{K}^{-1} \) (real cubic expansivity of mercury)
- \( \beta_g = 2.4 \times 10^{-5} \, \text{K}^{-1} \) (cubic expansivity of glass)
- \( \Delta T = 100^\circ C - 10^\circ C = 90 \, \text{K} \) (change in temperature)
First, we calculate \( \beta_m - \beta_g \):
\(\beta_m - \beta_g = 1.82 \times 10^{-4} - 2.4 \times 10^{-5} = 1.58 \times 10^{-4} \, \text{K}^{-1}\))
Now substitute the values into the apparent expansion formula:
\(\text{Apparent Expansion} = 100 \, \text{cm}^3 \cdot (1.58 \times 10^{-4}) \cdot 90\)
Calculating this gives:
\(= 100 \cdot 1.58 \times 10^{-4} \cdot 90\)
\(= 100 \cdot 1.422 \times 10^{-2}\)
\(= 1.422 \, \text{cm}^3\)
Thus, the apparent cubic expansion of the mercury is approximately:
\(\text{Answer: } 1.42 \, \text{cm}^3\)
Source: Myschool Past Questions Archive (Record #25069)
Physics
WAEC / WASSCE 1995
Thermal Physics
An iron rod of mass 2kg and at a temperature of 280oC is dropped into some quantity of water initially at a temperature of 30oC. lf the temperature of the mixture is 70oC, calculate the mass of the water. (Neglect heat losses to the surroundings). Specific heat capacity of iron = 460 J kgK) (Specific heat capacity of water = 4200 Jkg K)
A
0.58kg
B
0.77kg
C
1.15kg
D
1.50kg
E
2.04kg
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 #25070)
Physics
WAEC / WASSCE 1995
Thermal Physics
Solution
How much heat is required to convert 20 g of ice at 0oC to water at the same temperature? (Specific latent heat of ice = 336 J g-1)
A
1.35 x 103J
B
5.38 x 103J
C
6.72 x 103J
D
7.06 x 103J
E
8.06 x 103J
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The formula of specific latent heat of fusion is given by: Q = mL
: 20 X 336
= 6720 or 6.72 X 10\(^3\)
Source: Myschool Past Questions Archive (Record #25071)
Physics
WAEC / WASSCE 1995
Thermal Physics
Solution
When two objects P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that in Q. lf the masses of P and Q are the same, calculate the ratio of the specific heat capacities of Q to P.
A
4 : 1
B
2 : 1
C
1 : 1
D
1 : 2
E
1 : 4.
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The heat absorbed by a substance is given by the formula:
Q = m \times c \times \DeltaT
For objects P and Q.
- Same quantity of heat (Q) is supplied.
- Same mass (m).
- Temperature change in P is twice that in Q (\DeltaT\(_P\) = 2 \times \DeltaT\(_Q\)).
For P
Q = m \times c\(_P\) \times (2 \times \DeltaT\(_Q\))
For Q:
Q = m \times c\(_Q\) \times \DeltaT\(_Q\)
Since both equal **Q**, set them equal:
m \times c\(_P\) \times 2\DeltaT\(_Q\) = m \times c\(_Q\) \times \DeltaT\(_Q\)
Cancel m and \DeltaT\(_Q\) from both sides:
2 c\(_P\) = c\(_Q\)
Therefore:
\(\frac{c_Q }{ c_P}\) = 2
Ratio c\(_Q : c_P\) = 2 : 1
Source: Myschool Past Questions Archive (Record #25072)
Physics
UTME / JAMB 1995
Thermal Physics
The volume of an air bubble increases from the bottom to the top of a lake at constant temperature because
A
atmospheric pressure acts on the surface of the lake
B
pressure increases with depth
C
density remains constant with pressure
D
the bubble experiences an upthrust
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 #27780)
Physics
UTME / JAMB 1995
Thermal Physics
Temperature is the measure of the
A
quantity of heat transferred into the molecules of an object
B
mean kinetic energy of the molecules of the object
C
kinetic energy of any individual molecule of the substance
D
amount of work done by the molecules of the object
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 #27784)
Physics
UTME / JAMB 1995
Thermal Physics
Solution
A rectangular metal block of volume 10\(^{-6}\)m\(^3\) at 273k is heated to 573k. If its coefficient of linear expansion is 1.2 x 10\(^{-5}\)k\(^{-1}\), the percentage change of it volume is
A
1.5%
B
1.1%
C
0.4%
D
0.1%
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Y = \(\frac{\Delta V}{V \times \Delta \theta}\)
3 x 10-5 x 1.2 = \(\frac{\Delta\ V}{10^{-6}\times\ (573 - 273)}\)
\(\Delta\)V = 1.080 x 10\(^{-7}\)m\(^3\)
%\(\Delta\)V = \(\frac{1.080 \times\ 10^{-7}}{10^{-6}}\) x 100
= 1.08%
\(\approx\) = 1.1%
Source: Myschool Past Questions Archive (Record #27793)
Physics
UTME / JAMB 1995
Thermal Physics
A piece of wood floats inside water at room temperature with a fraction of it above the liquid surface. As the temperature of the water is raised, the part of the wood above the liquid will
A
decrease because the density of water decreases with temperature
B
increase because the density of water decreases with temperature
C
decreases because the density of water increases with water temperature
D
increase because the density of water increases with temperature
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 #27804)
Physics
UTME / JAMB 1995
Thermal Physics
Solution
A temperature scale has a lower fixed point of 40mm and a upper fixed point of 200mm. What is the reading on this scale when a thermometer reads 60oC?
A
33.3mm
B
36.0mm
C
96.0mm
D
136.0mm
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
200mm ..... 100oC
Xmm ..... 60oC
400mm ..... 0oC
\(\frac{x - 40}{200 - 40}\) = \(\frac{60 - 0}{100 - 0}\)
X = 136mm
Source: Myschool Past Questions Archive (Record #27815)