Thermal Energy and Phase Changes Quiz

Grade 9th

Reviewed by Ekaterina Yukhnovich

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By Thames

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Quizzes Created: 11119 | Total Attempts: 9,762,531

| Attempts: 12 | Questions: 20 | Updated: Mar 6, 2026

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1. If you accidentally record the hot object's initial temperature too high, the calculated specific heat of the metal will most likely be:

Too high

Too low

Unchanged

Exactly correct

Concept: error propagation in c=|q|/(m|Δt|). Recording t_i too high makes the calculated |Δt| too large.

Explanation

Concept: error propagation in c=|q|/(m|Δt|). Recording t_i too high makes the calculated |Δt| too large.

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About This Quiz

Thermal Energy and Phase Changes Quiz - Quiz

This quiz on thermal energy and phase changes includes 20 questions designed for students in Grade 9. You will explore key concepts such as calorimetry, heat transfer, and the behavior of substances during phase changes. Understanding these topics is important because they apply to real-world situations, like cooking and weathe... see morepatterns. By completing this quiz, you will strengthen your grasp of these principles and be better prepared for future science courses. Dive in and see how well you know the fascinating world of thermal energy!
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2. What first name or nickname would you like us to use?

You may optionally provide this to label your report, leaderboard, or certificate.

2. In calorimetry problems, which approach is usually the most reliable?

Use only q=mcΔt even during melting/boiling

Ignore signs and hope it works out

Break the process into stages and apply the correct equation to each stage

Always assume the final temperature is the average of the initial temperatures

Concept: stage-based problem solving. Calorimetry often involves different physical processes, each with its own equation.

Explanation

Concept: stage-based problem solving. Calorimetry often involves different physical processes, each with its own equation.

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3. Which could explain why a measured final temperature in a calorimetry lab is lower than the calculated insulated value?

Heat loss to the surrounding air

The cup absorbed some heat

Using a lid and better insulation

Hot object cooled a bit before being placed in the water

Concept: reasons real results are lower than ideal. Heat loss to air, heat absorbed by the cup, and cooling during transfer all reduce the energy that actually warms the water.

Explanation

Concept: reasons real results are lower than ideal. Heat loss to air, heat absorbed by the cup, and cooling during transfer all reduce the energy that actually warms the water.

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4. A metal sample loses 2,400 j while cooling by 20°C. If its mass is 0.30 kg, its specific heat is:

200 j/(kg·°C)

400 j/(kg·°C)

800 j/(kg·°C)

1,600 j/(kg·°C)

Concept: specific heat from q, m, and Δt. Use c=|q|/(m|Δt|). c=2400/(0.30×20)=400 j/(kg·°C).

Explanation

Concept: specific heat from q, m, and Δt. Use c=|q|/(m|Δt|). c=2400/(0.30×20)=400 j/(kg·°C).

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5. If 10,020 j is used to melt 0.030 kg of ice at 0°C, then the ice's latent heat of fusion is l_f=____ j/kg.

Concept: solving for latent heat. Use l_f=q/m for a melting process at constant temperature.

Explanation

Concept: solving for latent heat. Use l_f=q/m for a melting process at constant temperature.

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6. A 0.050 kg piece of ice at 0°C is dropped into warm water. If 16,700 j of heat is transferred to the ice, what happens to the ice (use l_f=334,000 j/kg)?

It warms to 10°C but stays solid

It melts completely

It partially melts only

It turns into steam

Concept: comparing transferred heat to melting requirement. Since the transferred heat exactly matches the value needed to melt all the ice, it melts completely.

Explanation

Concept: comparing transferred heat to melting requirement. Since the transferred heat exactly matches the value needed to melt all the ice, it melts completely.

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7. In a perfectly insulated mixing problem (no phase change), the final temperature must lie between the two initial temperatures.

Concept: equilibrium bounds in mixing. Without external energy, the mixture cannot end up hotter than the hottest part or colder than the coldest part.

Explanation

Concept: equilibrium bounds in mixing. Without external energy, the mixture cannot end up hotter than the hottest part or colder than the coldest part.

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8. A 0.20 kg sample of water warms from 18°C to 23°C. Heat absorbed is closest to:

836 j

4,180 j

8,360 j

20,900 j

Concept: applying q=mcΔt for warming. Here Δt=23−18=5°C. q=0.20×4180×5=4,180 j.

Explanation

Concept: applying q=mcΔt for warming. Here Δt=23−18=5°C. q=0.20×4180×5=4,180 j.

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9. Which statements about Δt are correct?

Δt=t_f-t_i

If an object cools, Δt is negative

If an object warms, Δt must be negative

Δt is always positive

Concept: temperature change sign. Δt is defined as final minus initial. Cooling makes t_f<t_i so Δt is negative.

Explanation

Concept: temperature change sign. Δt is defined as final minus initial. Cooling makes t_f<t_i so Δt is negative.

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10. If the system is not well insulated, the 'heat lost = heat gained' equation becomes less accurate.

Concept: non-isolated systems. If heat leaks to the surroundings, then not all the heat lost by the hot object ends up in the cold object.

Explanation

Concept: non-isolated systems. If heat leaks to the surroundings, then not all the heat lost by the hot object ends up in the cold object.

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11. When you add warm water to a cup of ice at 0°C (insulated), the first energy process is:

Water heats up

Ice warms above 0°C

Ice melts at 0°C using latent heat

Ice turns directly to steam

Concept: phase change priority at a melting point. Ice at 0°C cannot warm above 0°C until it has melted because added energy first goes into the phase change.

Explanation

Concept: phase change priority at a melting point. Ice at 0°C cannot warm above 0°C until it has melted because added energy first goes into the phase change.

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12. In the previous question (insulated), the metal's heat change is q_metal =____ j.

Concept: equal and opposite heat changes. In an insulated exchange, heat gained by water equals heat lost by the metal.

Explanation

Concept: equal and opposite heat changes. In an insulated exchange, heat gained by water equals heat lost by the metal.

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13. A 0.15 kg metal at 90°C is placed into 0.30 kg water at 20°C (ignore cup). If final temperature is 24°C, water gains:

502 j

5,016 j

12,540 j

5,016 j

Concept: heat gained by water. The water warms by Δt=24−20=4°C, so q=mcΔt=0.30×4180×4=5,016 j.

Explanation

Concept: heat gained by water. The water warms by Δt=24−20=4°C, so q=mcΔt=0.30×4180×4=5,016 j.

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14. In the previous question, the larger mass of water has more influence on the final temperature.

Concept: heat capacity (mc) determines influence. A larger mass has a larger heat capacity, so it stores more thermal energy per degree of temperature change.

Explanation

Concept: heat capacity (mc) determines influence. A larger mass has a larger heat capacity, so it stores more thermal energy per degree of temperature change.

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15. 0.40 kg of water at 30°C is mixed with 0.10 kg of water at 10°C (insulated). Final temperature is closest to:

14°C

18°C

26°C

40°C

Concept: mixing same substance. Use a mass-weighted average. t_f=(0.40(30)+0.10(10))/0.50=26°C.

Explanation

Concept: mixing same substance. Use a mass-weighted average. t_f=(0.40(30)+0.10(10))/0.50=26°C.

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16. Which steps might be needed to find final temperature when hot water meets ice (insulated)?

Compute heat hot water can release

Compute heat needed to melt ice

If all ice melts, compute warming of melted water

Ignore latent heat

Concept: multi-stage energy balance with phase change. You must compare heat available from cooling water with heat required to melt ice.

Explanation

Concept: multi-stage energy balance with phase change. You must compare heat available from cooling water with heat required to melt ice.

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17. A 0.25 kg sample of water cools from 45°C to 25°C. Heat released is closest to:

2,090 j

20,900 j

52,250 j

2,090 j

Concept: sensible heat release magnitude and sign. The magnitude is q=mc|Δt|=0.25×4180×20=20,900 j.

Explanation

Concept: sensible heat release magnitude and sign. The magnitude is q=mc|Δt|=0.25×4180×20=20,900 j.

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18. If there is still ice left in the cup, the final temperature must be 0°C (in an insulated system).

Concept: coexistence of phases fixes temperature. When ice and water coexist at equilibrium (at 1 atm), the temperature remains at 0°C.

Explanation

Concept: coexistence of phases fixes temperature. When ice and water coexist at equilibrium (at 1 atm), the temperature remains at 0°C.

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19. 0.030 kg of ice melts at 0°C. Using l_f=334,000 j/kg, the heat needed is:

1,002 j

10,020 j

100,200 j

11,140 j

Concept: applying q=ml_f. Multiply mass by latent heat of fusion. q=0.030×334,000=10,020 j, which is the energy required for complete melting at 0°C.

Explanation

Concept: applying q=ml_f. Multiply mass by latent heat of fusion. q=0.030×334,000=10,020 j, which is the energy required for complete melting at 0°C.

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20. The heat required to melt ice at 0°C is q=m____.

Concept: latent heat of fusion. Melting at constant temperature uses q=ml_f. l_f is the energy per kilogram required to change ice to liquid water at 0°C.

Explanation

Concept: latent heat of fusion. Melting at constant temperature uses q=ml_f. l_f is the energy per kilogram required to change ice to liquid water at 0°C.

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