A pot of soup, currently 66°C above room temperature, is left out to cool. If that temperature difference decreases by 10% per minute, then what will the difference be in 17 minutes?
We set up the temperature function T(m), where m is the number of minutes of cooling. With 10% = 0.1, we have:
T(m) = 66 * (1 - 0.10)^m
The problem asks for T(17) and the difference temperature:
T(17) = 66 * 0.9^17
T(17) = 66 * 0.16677181699
T(17) = 11.01C
Calculate the difference in temperature
Difference = Starting Temperature - Ending Temperature
Difference = 66 - 11.01
Difference = 66 - 11.01 = 54.99 ~ 55
We set up the temperature function T(m), where m is the number of minutes of cooling. With 10% = 0.1, we have:
T(m) = 66 * (1 - 0.10)^m
The problem asks for T(17) and the difference temperature:
T(17) = 66 * 0.9^17
T(17) = 66 * 0.16677181699
T(17) = 11.01C
Calculate the difference in temperature
Difference = Starting Temperature - Ending Temperature
Difference = 66 - 11.01
Difference = 66 - 11.01 = 54.99 ~ 55