Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
University Calculus, coffee cooling question.?
A cup of coffee at 180 degrees is poured into a mug and left in a room at 73 degrees. After 1 minutes, the coffee is 130 degrees. Assume that the differential equation describing Newton's Law of Cooling is (in this case) dT/dt= k(T-73).
What is the temperature of the coffee after 11 minutes?
After how many minutes will the coffee be 100 degrees?
2 Answers
- cidyahLv 78 years ago
Newton's Law of Cooling
T0 = Initial temperature
TA = Ambient temperature
t = time elapsed
T = Temperature after time t
T=TA+(T0-TA) e^kt
e^kt = (T-TA) / (T0-TA)
k = LN [ (T-TA) / (T0-TA)] / t
k = LN [ (130-73) / (180-73)] / 130
Constant of proportionality k = -0.629778
T = 73+107e^(-0.629778(t))
Find cooled temperature after time 11 minutes
T = 73+107e^(-0.629778)(t)
Substitute t = 11
T = 73+107e^(-0.629778)(11)
Cooled temperature is 73.105 degrees
When the temperature cools to 100:
t = LN [ (T-TA) / (T0-TA)] / k
t = LN [ (100-73) / (180-73)] / -0.629778
time = 2.1865 minutes
- jimmymae2000Lv 78 years ago
Cup will be 73 + e -t/1.414 (degrees F)
.
So after 11 min = 73.004 deg F
And after 1.414 minutes it will be 100 deg F
