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.

Question

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.

Answer 1

To find the substance's cooling rate, we need to use the data provided to solve for k in the equation T(t) = Ts + (T0 - Ts)e^(-kt).

Given:
T0 = 80°C
T(t) = 60°C
t = 15 min
Ts = 50°C

Substituting these values into the equation, we get:
60 = 50 + (80 - 50)e^(-k*15)

Simplifying the equation, we have:
10 = 30e^(-15k)

Divide both sides of the equation by 30:
1/3 = e^(-15k)

Take the natural logarithm (ln) of both sides:
ln(1/3) = -15k

Solve for k:
k = ln(1/3) / -15 ≈ 0.0389 (rounded to four decimal places)

Therefore, the substance's cooling rate when the surrounding air temperature is 50°C is approximately 0.0389.