A Journey through States of Water — Evidence, State Changes and Rates (2 of 3) · Answer guide

Lesson 1 · Test the source of outside droplets

Interpret whole-assembly mass readings and use comparisons without overstating the evidence.

1. What does the model’s 400.0 g starting reading include?

   Answer: Container, lid and all its contents on the balance. The measured assembly includes every part placed on the balance.

2. The same assembly increases from 400.0 g to 401.2 g. Its gain is…

   Answer: 1.2 g. 401.2 − 400.0 = 1.2 g.

3. Could water merely moving from inside to outside, while staying on the weighed assembly, explain a total mass gain by itself?

   Answer: No; that movement redistributes mass within the same measured system. Redistribution alone does not add matter to the assembly.

Discuss: Draw a boundary around everything on the balance. Calculate the supplied change and write a supported conclusion plus one remaining uncertainty.

Teaching note: Do not confuse total assembly mass with internal-water mass. “No change detected” does not always mean absolutely no change occurred, and a single observation should not be called conclusive proof of every cause. Open responses need discussion; they are not automatically marked.

Explain or demonstrate

For the supplied 520.0 g to 520.8 g closed-assembly record, calculate the change, identify the measured parts and explain why internal redistribution alone cannot cause the gain. Add one measurement limitation.

Evidence to include:

• System boundary named
• Mass gain calculated
• Conclusion and limitation separated

Review guide:

• 1 point: 0.8 g gain
• 1 point: lid/container/contents/outside droplets included
• 1 point: redistribution cannot add mass and finite sensitivity stated

Example response: The whole assembly gains 0.8 g. The reading includes the container, lid, contents and attached outside droplets. Moving water between its inside and outside cannot by itself add mass to that same assembly. A tiny level change could still be below the observation method’s resolution.

Lesson 2 · Follow changes of state

Name and correctly direct melting, freezing, evaporation and condensation.

1. Solid ice changing to liquid water is…

   Answer: Melting. Melting is solid to liquid.

2. Liquid water changing to solid ice is…

   Answer: Freezing. Freezing is liquid to solid.

3. Which arrow correctly describes condensation?

   Answer: Water vapour → liquid water. Condensation is gas to liquid.

Discuss: Draw a two-way solid–liquid–gas map with all four process names and arrows. Then describe a two-step pathway using the correct sequence.

Teaching note: Do not swap freezing with condensation: both can involve cooling, but their starting and ending states differ. A diagram arrow represents a process, not a piece of water physically turning into the word on the arrow. Open responses need discussion; they are not automatically marked.

Explain or demonstrate

Draw and label both directions between ice and liquid water, and both directions between liquid water and vapour. Explain one two-step pathway.

Evidence to include:

• Four arrows correctly directed
• Four names correctly matched
• Sequence explained

Review guide:

• 1 point: melting/freezing pair
• 1 point: evaporation/condensation pair
• 1 point: two-step sequence preserves state order

Example response: Ice → liquid is melting; liquid → ice is freezing. Liquid → vapour is evaporation; vapour → liquid is condensation. In the pathway ice → liquid → vapour, melting occurs before evaporation.

Lesson 3 · Compare evaporation rates fairly

Predict how conditions affect evaporation and design a comparison with one intended change.

1. With equal water amounts and comparable other conditions, a larger exposed area generally gives…

   Answer: Faster evaporation. More surface is exposed to the air.

2. To investigate exposed area, which should be kept the same?

   Answer: Starting water amount and surrounding conditions. Controls help link the result to the intended variable.

3. All else comparable, higher humidity usually makes evaporation…

   Answer: Slower. Moister surrounding air generally reduces the rate of evaporation.

Discuss: Plan one comparison for exposed area or air movement. Name the changed factor, three controls, the measured outcome and one possible limitation. Use supplied data if the setup is unavailable.

Teaching note: Bigger containers do not automatically prove a larger exposed water surface or equal starting amounts. Changing area, water amount and location together makes the comparison hard to interpret. Open responses need discussion; they are not automatically marked.

Explain or demonstrate

Plan a fair exposed-area comparison using equal water amounts. State the intended change, three controls, outcome and limitation.

Evidence to include:

• Area is the intended change
• Water amount/material/surroundings considered
• Clear timing rule and limit

Review guide:

• 1 point: changed factor clearly named
• 1 point: three relevant controls
• 1 point: measurable outcome and realistic limitation

Example response: I compare two exposed areas using the same starting water amount, comparable container material and the same surroundings. I record time until no visible liquid remains. Small remaining films may be difficult to see, so that observation rule has a limit.

Final check · Set 1

Try one set after the lessons. Use the other set after reviewing any errors.

1. Why use an initially dry room-temperature closed container as a comparison?

   To help examine the effect of a cold surface under similar surroundings. A comparison can help test the role of surface temperature.

2. A reliable mass gain and new outside droplets support which interpretation?

   Material arrived from outside the measured assembly. Condensation from surrounding air can add water to the weighed assembly.

3. An ice cube melts and the resulting water later evaporates. What is the order?

   Solid → liquid → gas. Follow the two stated processes in order.

4. Evaporation at room temperature obtains energy…

   From the water and its surrounding environment as energy is exchanged. Evaporation does not require a flame or boiling; energy exchanges occur with the surroundings.

5. Invented drying times are 42 minutes for a narrow surface and 18 for a wide one. Which dried first?

   The wide surface. 18 minutes is less than 42 minutes.

6. One sample starts with twice as much water and is placed in a different room. Can its drying-time difference isolate surface area alone?

   No; water amount and surroundings also changed. Several changed factors create alternative explanations.

Final check · Set 2

Try one set after the lessons. Use the other set after reviewing any errors.

1. A supplied assembly changes from 520.0 g to 520.8 g. What is the gain?

   0.8 g. 520.8 − 520.0 = 0.8 g.

2. Which report is most careful?

   The records support condensation from surrounding air; other small effects and measurement limits still need consideration. The conclusion follows the evidence while acknowledging limits.

3. Why are condensation and freezing different?

   They begin and end in different states. Condensation is gas-to-liquid; freezing is liquid-to-solid.

4. What should a state-change diagram label include?

   Starting state, ending state and correctly directed process. The process depends on direction as well as the named states.

5. Which is a clearly stated measured outcome?

   Time until no visible liquid remains under a stated observation rule. This describes an observable outcome with an explicit rule.

6. Why preserve an unexpected observation?

   It may reveal variation or an uncontrolled condition worth investigating. Unexpected results are data to examine, not discard silently.