Question

charlotte put two identical bottles of water in the freezer to cool. she put one bottle near the front of the freezer and one near the back. this table shows how the temperature of each bottle of water changed over 15 minutes.
bottle of water \tinitial temperature (°c) \tfinal temperature (°c)
bottle near the back \t20 \t11
bottle near the front \t21 \t15
the next time charlotte puts a bottle of water in the freezer, she wants the bottle to cool off quickly. what should she do to maximize the rate of thermal energy transfer out of the bottle of water?
put the bottle near the back
put the bottle near the front
either; the rate of thermal energy transfer is the same whether the bottle is near the back or the front

Explanation:

Step1: Calculate temp change (back)

$\Delta T_{\text{back}} = 20^\circ\text{C} - 11^\circ\text{C} = 9^\circ\text{C}$

Step2: Calculate temp change (front)

$\Delta T_{\text{front}} = 21^\circ\text{C} - 15^\circ\text{C} = 6^\circ\text{C}$

Step3: Compare cooling rates

The bottle near the back lost more thermal energy (a larger temperature drop) in the same 15-minute period, meaning it cooled faster.

Answer:

put the bottle near the back