Question 949100: Newton's Law of Cooling: A cup of hot chocolate has cooled from 92°C to 50°C after 12 min in a room at 22°C.
a) What is the value for k (round to the nearest ten-thousandth)?
b)How long will it take for the drink to cool to 30°C (Round to the nearest tenth of a minute and label your answer?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! The Newton's Law of Cooling Formula is given by
T(t) = Ts + (To - Ts)e^(-kt)
Where t is the time taken,
T(t) is the temperature of the given body at time t,
Ts is the surrounding temperature,
To is the initial temperature of the body,
k is the constant.
We have the following given and two problems to solve
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A cup of hot chocolate has cooled from 92°C to 50°C after 12 min in a room at 22°C.
a) What is the value for k (round to the nearest ten-thousandth)?
b)How long will it take for the drink to cool to 30°C (Round to the nearest tenth of a minute and label your answer?
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a) 50 = 22 + (92 - 22)e^(12k)
28 = 70e^(12k)
0.4 = e^(12k)
12k = ln(0.4)
k = −0.916290732 / 12 = −0.0763576
b) 30 = 22 + (92 - 22)e^(−0.0763576t)
8 = (92 - 22)e^(−0.0763576t)
0.114285714 = e^(−0.0763576t)
−0.0763576t = ln(0.114285714)
t = −2.169053703 / −0.0763576 = 28.40652015
The drink will cool to 30 degrees Celsius in 28.4 seconds
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