Heat Practice Questions

SPM (Malaysia) · SPM Physics · 145 free MCQs with instant results and detailed explanations.

145
Total
39
Easy
80
Medium
26
Hard

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Sample Questions from Heat

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Q1
Easy
What happens to the temperature of a substance when it undergoes a phase change, such as melting or boiling?
A. It increases continuously.
B. It decreases continuously.
C. It remains constant.
D. It fluctuates greatly.
Show Answer & Explanation
Correct Answer: C
During a phase change, such as melting or boiling, the temperature of a substance remains constant because all the heat energy is used in changing the state of the substance rather than increasing its temperature.
Q2
Easy
If a metal rod is heated at one end, how does the heat transfer through the rod occur?
A. Conduction
B. Convection
C. Radiation
D. Evaporation
Show Answer & Explanation
Correct Answer: A
Heat transfer through a solid object like a metal rod occurs by conduction, where kinetic energy is transferred from molecule to molecule through direct contact.
Q3
Easy
Which of the following materials is the best insulator of heat?
A. Copper
B. Wood
C. Glass
D. Air
Show Answer & Explanation
Correct Answer: D
Air is considered the best insulator of heat among the given options because it has low thermal conductivity, preventing heat transfer effectively.
Q4
Medium
What is the main principle behind a thermos flask's design that allows it to keep drinks hot or cold?
A. Conduction
B. Convection
C. Radiation
D. Insulation
Show Answer & Explanation
Correct Answer: D
The thermos flask is designed to minimize heat transfer through insulation, preventing heat loss or gain from external environments.
Q5
Medium
A metal rod of length 2 m expands by 0.004 m when heated. What is the coefficient of linear expansion for the metal?
A. 2 x 10^-5 /ยฐC
B. 1 x 10^-5 /ยฐC
C. 4 x 10^-5 /ยฐC
D. 3 x 10^-5 /ยฐC
Show Answer & Explanation
Correct Answer: C
The coefficient of linear expansion (ฮฑ) is calculated using the formula ฮฑ = ฮ”L / (L0 * ฮ”T). The expansion leads to a coefficient of 4 x 10^-5 /ยฐC.
Q6
Medium
A 500 g block of ice at -10 ยฐC is placed in 1 kg of water at 20 ยฐC. Assuming no heat loss to the surroundings, what is the final temperature of the mixture when the ice melts completely?
A. 0 ยฐC
B. 10 ยฐC
C. 5 ยฐC
D. 15 ยฐC
Show Answer & Explanation
Correct Answer: A
The heat gained by the melting ice equals the heat lost by the water, leading to a final equilibrium temperature of 0 ยฐC.
Q7
Medium
Which of the following factors does NOT affect the rate of heat transfer by conduction?
A. Material of the conductor
B. Temperature difference
C. Surface area
D. Time of exposure
Show Answer & Explanation
Correct Answer: D
The rate of heat transfer by conduction depends on material, temperature difference, and surface area, but not on the time of exposure.
Q8
Hard
Consider a container with 2 kg of water at 50ยฐC mixed with 1 kg of ice at 0ยฐC. What will be the final temperature of the mixture after thermal equilibrium is reached? (Assume no heat loss to the environment and specific heat capacity of water is 4.2 kJ/kgยฐC)
A. 10ยฐC
B. 20ยฐC
C. 30ยฐC
D. 40ยฐC
Show Answer & Explanation
Correct Answer: C
To find the final temperature, we set up the heat lost by the water equal to the heat gained by the ice. The water cools down while the ice melts and warms up. The heat lost by the water can be calculated using Q = mcฮ”T, and the heat gained by the ice includes the latent heat of fusion. The calculations show the final equilibrium temperature to be 30ยฐC.
Q9
Hard
A block of ice at 0ยฐC is placed in a calorimeter containing 200 g of water at 25ยฐC. If the specific heat capacity of water is 4.18 J/gยฐC and the latent heat of fusion for ice is 334 J/g, what is the final temperature of the system when thermal equilibrium is reached? Assume no heat is lost to the surroundings.
A. 0ยฐC
B. 25ยฐC
C. 15ยฐC
D. 10ยฐC
Show Answer & Explanation
Correct Answer: D
The final temperature is 10ยฐC because the heat lost by the water is equal to the heat gained by the ice during melting and warming up. Calculating the heat lost by 200 g of water cooling from 25ยฐC to 10ยฐC and matching it to the heat gained by melting the ice and warming it to the same temperature confirms this.
Q10
Hard
A 1500 W electric heater is used to heat a metal rod of mass 2 kg. If the specific heat capacity of the metal is 0.385 J/gยฐC, how long will it take to raise the temperature of the metal rod from 20ยฐC to 100ยฐC?
A. 40 minutes
B. 30 minutes
C. 20 minutes
D. 10 minutes
Show Answer & Explanation
Correct Answer: B
It takes 30 minutes to raise the temperature because first we calculate the total heat required using Q = mcฮ”T, then divide that by the power of the heater to find the time taken.

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Heat โ€” SPM (Malaysia) SPM Physics Practice Questions Online

This page contains 145 practice MCQs for the chapter Heat in SPM (Malaysia) SPM Physics. The questions are organized by difficulty โ€” 39 easy, 80 medium, 26 hard โ€” so you can choose the right level for your preparation.

Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit.