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Answer :
To the nearest minute, it will take approximately 30 minutes for the soup to cool from 100°F to 80°F.
The rate at which the soup cools can be approximated using Newton's law of cooling. According to the law, the rate of cooling is proportional to the temperature difference between the object and its surroundings. In this case, the initial temperature of the soup is 100°F, and the room temperature is 68°F, resulting in a temperature difference of 32°F.
We know that after 20 minutes, the soup's temperature dropped to 95°F. From this information, we can calculate the rate of cooling. In 20 minutes, the soup cooled by 5°F, indicating a cooling rate of 5°F/20 minutes or 0.25°F/minute.
To find the time it takes for the soup to cool to 80°F, we need to determine how many minutes it takes for the temperature to drop from 95°F to 80°F. This temperature difference is 15°F. Dividing the temperature difference by the cooling rate gives us 15°F / 0.25°F/minute = 60 minutes.
Therefore, it will take approximately 60 minutes for the soup to cool from 95°F to 80°F. Adding the initial 20 minutes, the total time required for the soup to cool from 100°F to 80°F is approximately 80 minutes.
To learn more about Newton's law of cooling click here:
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