Interpret warm and cold temperature correction data
Use the supplied reference, departure magnitudes and matched time intervals.
Extended only (Supplement); Biology 0610, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked
Before you begin
- Recall hormone production and blood transport, respiration, liver storage and electrical impulses along neurones.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is each departure magnitude after five minutes?
What you will learn
- Read the invented temperature records
- Calculate signed rates and limited conclusions
Interpret warm and cold temperature correction data
Invented reference 37.0°C: warm 38.0→37.4→37.0; cold 36.0→36.6→37.0 at 0, 5 and 10 min. Opposite signs, same shrinking departure. Original teaching model.
| Time / min | Warm start / °C | Cold start / °C |
|---|---|---|
| 0 min | 38.0 | 36.0 |
| 5 min | 37.4 | 36.6 |
| 10 min | 37.0 | 37.0 |
| Feature | Meaning |
|---|---|
| Departures | 1.0→0.4→0°C for both |
| First interval | Warm −0.12; cold +0.12°C/min |
| Second interval | Warm −0.08; cold +0.08°C/min |
| Limits | Mean rates; no sole-effector proof or universal exact temperature |
Read the invented temperature records
The invented model supplies a reference of 37.0°C. At 0, 5 and 10 minutes, the warm-start temperatures are 38.0, 37.4 and 37.0°C, while the cold-start values are 36.0, 36.6 and 37.0°C. Both departure magnitudes shrink from 1.0 to 0.4 to 0°C. The reference belongs to this model rather than being an exact universal value for everyone.
For the first five minutes, warm signed change is 37.4 − 38.0 = −0.6°C, giving mean rate −0.6/5 = −0.12°C per minute. Cold signed change is +0.6°C, giving +0.12°C per minute. Both move towards the supplied reference despite opposite signs. If reporting a cooling magnitude, use 0.12°C per minute and explicitly name cooling.
Calculate signed rates and limited conclusions
Over the second five-minute interval, the warm change is −0.4°C and the cold change is +0.4°C, giving mean rates −0.08 and +0.08°C per minute. These are interval averages, not instantaneous derivatives or direct sweat-production measurements. Always identify the interval before comparing rates.
The warm pattern is consistent with coordinated heat-loss responses and the cold pattern with increased heat production and conservation. Temperature readings alone do not establish which exact effector was solely responsible. The data are for explanation and calculation, not a clinical threshold or an instruction to heat or chill human subjects.
Pause and explain
Which second-interval rate pair is correct?
Worked example
Calculate first-interval rates and compare correction.
Show the worked solution
- Warm: −0.6/5 = −0.12°C/min.
- Cold: +0.6/5 = +0.12°C/min.
Answer First-interval mean rates: warm −0.12°C/min; cold +0.12°C/min. Both reduce departure from reference 37.0°C.
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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.
Calculate departure magnitudes for both records at each displayed time.
Give me a hint
Use distance from 37.0°C.
Compare my reasoning
- At zero minutes both differ by 1.0°C.
- At five minutes by 0.4°C; at ten by zero.
Both records have departure magnitudes 1.0, 0.4 and 0°C at 0, 5 and 10 minutes.
Look for these in your work
- Use the supplied reference
- Compare matching times
Calculate signed mean rates for both five-minute intervals in both records.
Give me a hint
Later minus earlier, divided by five.
Compare my reasoning
- First: warm −0.12; cold +0.12°C/min.
- Second: warm −0.08; cold +0.08°C/min.
First interval: warm −0.12, cold +0.12°C/min. Second interval: warm −0.08, cold +0.08°C/min.
Look for these in your work
- Calculate both intervals
- Retain rate units and signs
Evaluate: these records prove sweating was the sole warm response and 37.0°C is exact for all people.
Give me a hint
Separate model consistency and universal claims.
Compare my reasoning
- No direct effector or controlled causal measurement is supplied.
- 37.0°C is the model reference, not a universal exact reading.
The record is consistent with correction, but does not prove sweating was the sole cause. The supplied 37.0°C reference is not an exact universal temperature for all people.
Look for these in your work
- Limit causal inference
- Keep the reference model-specific
Common mistakes
- Calling every negative signed rate failed negative feedback.
- Treating interval averages as instantaneous or clinical values.
What can you explain now?
Which second-interval rate pair is correct?
Compare with the explanation
Warm −0.08; cold +0.08°C/min
Each changes 0.4°C in five minutes.
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Make it stick
Calculate departures and rates while retaining the limits of a temperature-only record.
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.