41. Cooling soup Suppose that a cup of soup cooled from 90°C to 60°C after 10 min in a room where the temperature was 20°C. Use Newton’s Law of Cooling to answer the following questions.
a. How much longer would it take the soup to cool to 35°C?

Verified step by step guidance

1. Recall Newton's Law of Cooling, which states that the temperature of an object changes at a rate proportional to the difference between its temperature and the ambient temperature. Mathematically, this is expressed as: \[ T(t) = T_{ambient} + (T_0 - T_{ambient}) e^{-kt} \] where \[T(t)\] is the temperature at time \[t\], \[T_0\] is the initial temperature, \[T_{ambient}\] is the room temperature, and \[k\] is a positive constant.
2. Identify the known values from the problem: initial temperature \[T_0 = 90^\circ C\], ambient temperature \[T_{ambient} = 20^\circ C\], temperature after 10 minutes \[T(10) = 60^\circ C\]. Substitute these into the formula to find the constant \[k\]: \[ 60 = 20 + (90 - 20) e^{-10k} \].
3. Solve the equation for \[k\] by isolating the exponential term: \[ 60 - 20 = 70 e^{-10k} \], which simplifies to \[ 40 = 70 e^{-10k} \]. Then, divide both sides by 70 and take the natural logarithm to solve for \[k\]: \[ e^{-10k} = \frac{40}{70} \], so \[ -10k = \ln\left(\frac{40}{70}\right) \], and finally \[ k = -\frac{1}{10} \ln\left(\frac{40}{70}\right) \].
4. Use the value of \[k\] to find the time \[t\] when the soup cools to \[35^\circ C\]. Substitute \[T(t) = 35\] into the original formula: \[ 35 = 20 + 70 e^{-kt} \]. Isolate the exponential term: \[ 15 = 70 e^{-kt} \], then \[ e^{-kt} = \frac{15}{70} \].
5. Solve for \[t\] by taking the natural logarithm: \[ -kt = \ln\left(\frac{15}{70}\right) \], so \[ t = -\frac{1}{k} \ln\left(\frac{15}{70}\right) \]. Finally, subtract the initial 10 minutes to find how much longer it takes to cool from 60°C to 35°C: \[ t_{additional} = t - 10 \]. This gives the additional cooling time needed.

Key Concepts

Newton’s Law of Cooling

Newton’s Law of Cooling states that the rate of change of an object's temperature is proportional to the difference between its temperature and the ambient temperature. Mathematically, it is expressed as dT/dt = -k(T - T_env), where k is a positive constant. This law models how objects cool or warm over time toward the surrounding temperature.

Solving First-Order Differential Equations

The cooling process is described by a first-order linear differential equation. Solving it involves separating variables or using integrating factors to find a function T(t) that gives temperature at time t. This solution typically has the form T(t) = T_env + (T_initial - T_env)e^(-kt), which helps predict future temperatures.

Exponential Decay and Time Calculation

The temperature difference decreases exponentially over time, meaning the soup cools rapidly at first and then more slowly. To find the time to reach a specific temperature, you solve for t in the exponential decay formula. This requires using logarithms to isolate t and calculate the additional cooling time accurately.