Explain cooling of an object touching an evaporating liquid
Trace energy from a contacted object into an evaporating liquid and account for replenishment.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Explain evaporation cooling of the liquid.
- Distinguish energy transfer direction from the movement of cold.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which energy direction explains the contacted object's cooling?
What you will learn
- Explain cooling of an object in contact with an evaporating liquid.
- Distinguish a cooling tendency from a guaranteed temperature fall when energy is replenished.
Follow energy across the contact
In a specified wet-cloth-and-object model, liquid evaporates from the exposed cloth and requires energy. Energy can transfer from the contacted object into the liquid, reducing the object's internal energy and cooling it when that energy is not fully replaced. The cloth provides a liquid in thermal contact; the explanation is not that cold flows into the object.
Trace the chain: object supplies energy to contacted liquid; evaporation removes energy and material from the liquid boundary. Surroundings may also supply energy. The schematic arrow directions identify transfers in this selected account rather than measured rates, temperatures or a certified device. It is a physics teaching model, not a medical cooling recommendation.
Balance all relevant transfers
If a supplied model removes 12 J/s from the object through this route and other transfers provide 5 J/s to the object, the net internal-energy loss is 7 J/s. Over 60 s the object loses 420 J if the rates stay constant and no other destinations apply. These rates are supplied account information; no phase-change energy constant or temperature drop is calculated from them alone.
When energy supplied to the object matches the loss, its temperature can remain steady within the stated same-state account. Reduced evaporation or different surrounding conditions can alter the effect. Do not claim every wet object must cool by the same number of degrees, or that contact alone establishes the energy route without the stated evaporation and transfer conditions.
Sources for this lesson
Pause and explain
Loss through the stated route is 12 W and replacement is 5 W. What is the net loss rate?
Worked example
A supplied object account loses energy at 12 W through evaporation at a contacted wet layer and receives 5 W from other transfers. Find net loss over 60 s; rates are constant and no other transfers apply.
Show the worked solution
- Net energy-loss rate is 12 − 5 = 7 W.
- Energy lost is 7 × 60 = 420 J.
- An exact temperature drop needs the object's thermal properties and state assumptions.
Answer 420 J net loss; the exact temperature drop is not determined.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Explain how a wet layer in contact with an object can cool the object as liquid evaporates.
Give me a hint
Trace the energy route and the replacement condition.
Compare my reasoning
- Evaporation requires energy at the liquid surface.
- The contacted object can supply energy into the liquid.
- Without full replacement the object loses internal energy and can cool.
Energy passes from the object to the evaporating liquid, producing cooling without full replenishment.
Look for these in your work
- Name the energy direction.
- Keep evaporation and compensation explicit.
A supplied object loses 12 W by the contacted evaporation route and receives 5 W otherwise, for 60 s. Calculate net signed energy change.
Give me a hint
Use the object boundary and subtract loss from input.
Compare my reasoning
- Net rate into the object is 5 − 12 = −7 W.
- ΔE = −7 × 60 = −420 J.
- The positive amount lost is 420 J.
ΔE = −420 J; 420 J lost.
Look for these in your work
- Preserve the named-object sign.
- Retain constant-rate and no-other-transfer assumptions.
A wet contacted object stays at steady temperature while evaporating liquid receives compensating energy. Does this disprove evaporation cooling? Explain why no exact drop follows.
Give me a hint
The complete energy account may compensate for the cooling tendency.
Compare my reasoning
- Evaporation can still occur and require energy.
- Other transfers can replace energy that would otherwise cool the object.
- An exact drop needs unbalanced sample energy and thermal properties; the steady account supplies no such drop.
No; compensation can keep temperature steady despite the cooling tendency.
Look for these in your work
- Account for all stated transfers.
- Avoid a universal temperature-drop claim.
Common mistakes
- Energy transfers out; cold is not a substance moving in.
- Replenishment can compensate for the cooling effect.
- A supplied rate is not a universal evaporation constant.
What can you explain now?
Why can a contacted object cool while liquid evaporates from it?
Compare with the explanation
It supplies energy to the evaporating liquid.
Without full replacement, that transfer reduces the object's internal energy.
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Make it stick
Trace contacted-object energy loss and show how replenishment changes the observed temperature outcome.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.