What should Shelley do to minimize the rate of thermal energy transfer out of the hot chocolate?

Understand the Problem

The question is asking what action Shelley should take to minimize the cooling of the hot chocolate in outdoor conditions. This involves understanding thermal energy transfer and how the size of the thermos might impact that rate.

Answer

Put the hot chocolate in the smaller thermos.

Steps to Solve

Analyze Temperature Changes

From the table, we see the temperature changes:
- Smaller thermos: Initial $95^\circ C$, Final $69^\circ C$
- Larger thermos: Initial $92^\circ C$, Final $61^\circ C$
Determine Cooling Rate

Calculate the change in temperature for each thermos over the 2-hour period.
- Smaller thermos: $$ \Delta T = 95 - 69 = 26^\circ C $$
- Larger thermos: $$ \Delta T = 92 - 61 = 31^\circ C $$
Relate Thermos Size to Cooling

A larger thermos has a greater surface area, which can lead to a higher rate of thermal energy transfer. The observations suggest that the larger thermos (with a greater temperature drop) cools faster than the smaller thermos.
Conclusion about Thermos Selection

To minimize cooling, Shelley should choose the thermos that cools less quickly. The smaller thermos exhibited a smaller temperature drop, suggesting it retains heat better under cold conditions.

Shelley should put the hot chocolate in the smaller thermos.

More Information

The smaller thermos, despite containing the same amount of hot chocolate, results in less thermal energy loss due to its lower surface area exposure, which helps retain heat for a longer duration.

Tips

Assuming that the amount of hot chocolate in each thermos is the sole factor; the size and surface area also play crucial roles in thermal energy transfer.
Not considering the initial and final temperatures correctly when determining which thermos was more effective at retaining heat.

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