Question

on a cold day, lucy brought two thermoses of hot chocolate outside. the amount of hot chocolate in each thermos was the same, but one thermos was larger than the other. this table shows how the temperature of each hot chocolate changed over 2 hours.

hot chocolate	initial temperature (°c)	final temperature (°c)
hot chocolate in the smaller thermos	94	62

the next time lucy brings hot chocolate outside, she does not want it to cool down quickly. what should she do to minimize the rate of thermal energy transfer out of the hot chocolate?
put the hot chocolate in the larger thermos
put the hot chocolate in the smaller thermos
either; the rate of thermal energy transfer is the same whether the hot chocolate is in the smaller or larger thermos

Explanation:

Step1: Analyze temperature - change data

The hot chocolate in the larger thermos went from 91°C to 53°C (a change of 91 - 53=38°C). The hot chocolate in the smaller thermos went from 94°C to 62°C (a change of 94 - 62 = 32°C) in 2 hours. But we need to consider the concept of surface - area to volume ratio. A larger volume (larger thermos) relative to surface area loses heat more slowly.

Step2: Recall heat - transfer principle

The rate of heat transfer is related to the surface - area to volume ratio. A larger object (larger thermos) has a smaller surface - area to volume ratio compared to a smaller object (smaller thermos) when the amount of substance is the same. A smaller surface - area to volume ratio means less heat is transferred per unit time.

Answer:

put the hot chocolate in the larger thermos