Newton’s law of cooling states that for a cooling substance with initial temperature T0, the temperature T(t) after t minutes can be modeled by the equation T(t)=Ts+(T0−Ts)e−kt, where Ts is the surrounding temperature and k is the substance’s cooling rate.
A liquid substance is heated to 80°C. Upon being removed from the heat, it cools to 60°C in 15 min.
What is the substance’s cooling rate when the surrounding air temperature is 50°C?
Round the answer to four decimal places.
0.0687
0.0732
0.0813
0.0872
I guessed 0.0813 which was wrong. The answer is 0.0732
It's 0.0916
so why did you guess? They gave you the formula and the data point.
T(15) = 60, so
50 + 30e^(-15k) = 60
k = ln3/15 = 0.0732
To find the substance's cooling rate, we need to use the given information and the equation for Newton's law of cooling.
The equation for Newton's law of cooling is: T(t) = Ts + (T0 - Ts) * e^(-kt)
Given information:
Initial temperature (T0) = 80°C
Temperature after 15 minutes (T(15)) = 60°C
Surrounding temperature (Ts) = 50°C
We need to use these values to find the cooling rate (k).
First, let's substitute the given values into the equation:
60 = 50 + (80 - 50) * e^(-k * 15)
Simplifying further:
10 = 30 * e^(-15k)
Dividing both sides of the equation by 30:
1/3 = e^(-15k)
To isolate the exponent, we can take the natural logarithm (ln) of both sides:
ln(1/3) = ln(e^(-15k))
Using the property of logarithms, ln(e^x) = x:
ln(1/3) = (-15k)
Now, we can solve for k by dividing both sides by -15:
k = ln(1/3) / (-15)
Using a calculator, we find:
k ≈ 0.0872 (rounded to four decimal places)
Therefore, the substance's cooling rate when the surrounding air temperature is 50°C is approximately 0.0872.