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Physics
UTME / JAMB 1998
Thermal Physics
Solution
How many grams of water at 17oC must be added to 42g of ice at 0oC to melt the ice completely? [specific latent heat of fusion of ice = 3.4 x 105Jkg-1, specific heat capacity of water = 4200Jkg-1k-1]
A
200g
B
300g
C
320g
D
400g
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Heat lost = Heat gained
mc\(\theta\) = mL
m x 4200 x 17 = 42 x 3.4 x 105
therefore, m = 200g
Source: Myschool Past Questions Archive (Record #27634)
Physics
UTME / JAMB 1998
Thermal Physics
Which of the following are true about the infra-red radiation? i. it is invisible ii. it is called heat ray iii. its frequency is higher than that of blue light iv. it travels as a transverse wave
A
i, ii, iii and iv
B
i, ii, iv only
C
i, iii, iv only
D
ii, iii, iv only
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 #27658)
Physics
WAEC / WASSCE 1998
Thermal Physics
Solution
The lower and upper fixed points of a mecury-in-glass thermometer are marked Y and 180mm respectively. On a particular the mecury meniscus in the thermometer rises to 45mm. If the corresponding reading on a scale is 10o, calculate the value of Y.
A
4mm
B
18mm
C
30mm
D
135mm
E
450mm
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
by interpolation,
\(\frac{100 - 0}{10- 0} = \frac{180 - y}{45 - y}\)
\(\frac{100}{10} = \frac{180 -y}{45 - y }\)
450 - 10y = 180 - y
- 9y = - 270
y = 30mm
Source: Myschool Past Questions Archive (Record #33552)
Physics
WAEC / WASSCE 1998
Thermal Physics
Solution
How much heat is emitted when a body of mass 200g cools from 37oC to 31oC? (specific heat capacity of the body = 0.4Jg-1K-1,)
A
4800J
B
1200J
C
480J
D
202J
E
80J
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
H = mC\(\theta\) = 200 x 0.4 x 6
= 480J
Source: Myschool Past Questions Archive (Record #33564)
Physics
WAEC / WASSCE 1997
Thermal Physics
Solution
A solid material of volume 100\(cm^{-3}\) is heated through a temperature difference of 40°C. Calculate the increase in the volume of the material if its linear expansitivity is 2 x \(10^{-6}K{-1}\)
A
2.4 x\(10^{-2}cm^{-3}\)
B
1.6 x\(10^{-2}cm^{-3}\)
C
8 x\(10^{-2}cm^{-3}\)
D
5 x\(10^{-2}cm^{-3}\)
E
4 x\(10^{-2}cm^{-3}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
\(3{\alpha}= \frac{\Delta{V}}{V_1\Delta{T}}\Rightarrow \frac{\Delta{V}}{100\times40}=3\alpha\, where\, \alpha = 2 \times 10^{-6}\\ \frac{\Delta{V}}{4000}=6\times10^{-6}\Rightarrow \Delta=4\times10^3\times6\times10^{-6}= 2.4\times10^{-2}cm^{-3}\)
Source: Myschool Past Questions Archive (Record #26301)
Physics
WAEC / WASSCE 1997
Thermal Physics
Solution
Calculate the quantity of heat released when 100g of steam at 100°C condenses to water(Take the specific latent heat of vaporization of water as 2.3x \(10^6Jkg^{-1}\))
A
2.3x\(10^{1}J\)
B
2.3x\(10^{2}J\)
C
2.3x\(10^{4}J\)
D
2.3x\(10^{5}J\)
E
3.3x\(10^{7}J\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
\(\Delta{H}=mL=0.1\times2.3\times10^6=2.3\times10^5\)J
Source: Myschool Past Questions Archive (Record #26302)
Physics
WAEC / WASSCE 1997
Thermal Physics
Solution
A cup containing 100g of pure water at 20°C is placed in a refrigerator. If the refrigerator extracts heat at a rate of 840J per minute, calculate the time taken for the water to freeze. Neglect the heat capacity of the material of the cup. (Specific heat capacity of water =4.2J\(g^{-1}K^{-2}\))((Specific latent heat fusion of water =336J\(g^{-1}\)
A
15minutes
B
20 minutes
C
42 minutes
D
50 minutes
E
84 minutes
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
H=pt =mL +mcθ ; 840t= 100x336+100x4.2x20 thus t=420000/840=50minutes
Source: Myschool Past Questions Archive (Record #26305)
Physics
UTME / JAMB 1997
Thermal Physics
Solution
The length of mercury thread when it is at 0^\circC, 100^\circC and at an unknown temperature \(\theta\) is 25mm, 225mm and 175mm respectively. The value of \(\theta\) is
A
85.0oC
B
80.0oC
C
75.0oC
D
70.0oC
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(\frac{175 - 25}{225 - 25} = \frac{x - 0}{100 - 0}\)
\(\frac{150}{200} = \frac{x}{100}\)
\(x = \frac{150 \times 100}{200}\)
= 75^\circC
Source: Myschool Past Questions Archive (Record #27692)
Physics
UTME / JAMB 1997
Thermal Physics
Equal masses of copper and rubber are raised to the same temperature. After sometimes, the copper was observed to be at a lower temperature because
A
the specific heat capacity of copper is lower than that of rubber
B
copper expands more than rubber
C
the specific heat capacity of rubber is lower than that of copper
D
rubber expands more than copper
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 #27693)
Physics
UTME / JAMB 1997
Thermal Physics
Solution
A mass of a liquid at 30oC is mixed with a mass of the same liquid at 70oC and the temperature of the mixture is 45oC. Find the ratio of the mass of the cold liquid to the mass of the other liquid.
A
3 : 5
B
5 : 3
C
3 : 7
D
7 : 3
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Heat lost = heat gained
mH x c x (70 - 45) = mc x c x (45 - 30)
\(\frac{m_c}{m_H}\) = \(\frac{5}{3}\)
Source: Myschool Past Questions Archive (Record #27697)
Physics
UTME / JAMB 1997
Thermal Physics
What happens when a gas expands at constant temperature?
A
its pressure decreases
B
the total momentum of its molecules remains constant
C
its pressure decreases and the total momentum of its moleculees remains constant
D
its pressure decreses and the total kinetic energy of its molecule decreases
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 #27698)
Physics
UTME / JAMB 1997
Thermal Physics
The electromagnetic waves that are sensitive to temperature changes are
A
X-rays
B
Gamma-rays
C
Ultra-violet rays
D
Infra-red rays
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 #27715)
Physics
WAEC / WASSCE 1996
Thermal Physics
Solution
A platinum-resistance thermometer has a resistance of 5\Omega at 0^\circC and 9\Omega at l00^\circC. Assuming that resistance changes uniformly with temperature, calculate the resistance of the thermometer when the temperature is 45^\circC.
A
13.9 Ω
B
8.9 Ω
C
6.8 Ω
D
3.2 Ω
E
1.8 Ω
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(\frac{\theta}{100} = \frac{R_ {\theta} - R_0}{R_{100} - R_0}\)
\(\frac{45}{100} = \frac{R_{\theta} - 5}{9 - 5}\)
\(45 \times 4 = 100(R_{\theta} - 5)\)
\(180 = 100R_{\theta} - 500\)
\(180 + 500 = 100R_{\theta}\)
\(R_{\theta} = \frac{680}{100}\)
\(R_{\theta} = 6.8 \Omega\)
Source: Myschool Past Questions Archive (Record #25210)
Physics
WAEC / WASSCE 1996
Thermal Physics
The linear expansivity of a metal P is twice that of another metal Q. When these materials are heated through the same temperature change, their increase in length is the same. Calculate the ratio of the original length of P to that of Q.
A
1:4
B
1:2
C
2:1
D
4:1
E
6:1
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 #25211)
Physics
WAEC / WASSCE 1996
Thermal Physics
A 400W immersion heater is used to heat a liquid of mass 0.5kg. lf the temperature of the liquid increases by 2.5oC in one second, calculate the specific heat capacity of the liquid (Neglect heat losses to the surroundings)
A
80 Jkg-1K-1
B
320Jkg-1K-1
C
500Jkg-1K-1
D
1200J kg-1K-1
E
2000J kg-1K-1.
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 #25214)