Question

The rate at which liquids cool can be modelled by an approximation of Newton's law of cooling, \(T(t)=\left(T_{i}-T_{f}\right)(0.9)^{\frac{t}{5}}+T_{f},\) where \(T_{f}\) represents the final temperature, in degrees Celsius; \(T_{i}\) represents the initial temperature, in degrees Celsius; and \(t\) represents the elapsed time, in minutes. Suppose a cup of coffee is at an initial temperature of \(95^{\circ} \mathrm{C}\) and cools to a temperature of \(20^{\circ} \mathrm{C}\). a) State the parameters \(a, b, h,\) and \(k\) for this situation. Describe the transformation that corresponds to each parameter. b) Sketch a graph showing the temperature of the coffee over a period of 200 min. c) What is the approximate temperature of the coffee after 100 min? d) What does the horizontal asymptote of the graph represent?

Short answer

a) \(a = 75, b = 0.9, h = 0, k = 20\)b) Graph needs to be plotted from 95°C to 20°C over 200 minutes.c) \(T(100) \approx 29.12^{\circ}C\)d) The horizontal asymptote is 20°C representing the final temperature.

Step-by-step solution

Understanding the Problem

Given the function for the cooling rate of a liquid: \[T(t)=\left(T_{i}-T_{f}\right)(0.9)^{\frac{t}{5}}+T_{f},\] it describes the temperature over time, where \(T_{i}\) is the initial temperature, \(T_{f}\) is the final temperature, and \(t\) is the time in minutes.

Part A: Identifying Parameters

In this context: \(T_{i} = 95^{\circ}C\) is the initial temperature,\(T_{f} = 20^{\circ}C\) is the final temperature.Identify the parameters 'a', 'b', 'h', and 'k' as follows:- \(a\): The factor that determines the initial effect of temperature change, i.e., \(T_i - T_f = 95 - 20 = 75\).- \(b\): Exponential base, which is 0.9.- \(h\): Horizontal shift (time) is 0 (since there's no shift in this model).- \(k\): Final temperature, \(T_{f} = 20^{\circ}C\).Thus, the parameters are \[a = 75, b = 0.9, h = 0, k = 20.\]

Part B: Sketching the Graph

To sketch the graph: Start at the initial temperature and plot points at various times using the formula: \(T(t)=75\cdot (0.9)^{\frac{t}{5}}+20\). Calculate values for \(t = 0, 50, 100, 150, 200\). Then draw the curve starting from 95°C approaching 20°C as time increases.

Part C: Calculating Temperature After 100 Minutes

Substitute \(t = 100\) min in the formula:\[T(100) = 75\cdot(0.9)^{\frac{100}{5}} + 20.\] Simplify the exponent: \[T(100) = 75\cdot(0.9)^{20}+20.\]Use a calculator for precise value: \(0.9^{20} \approx 0.121577.\)Thus: \[ T(100) \approx 75 \cdot 0.121577 + 20 \approx 9.12 + 20 \approx 29.12^{\circ}C.\]

Part D: Horizontal Asymptote

The horizontal asymptote represents the final temperature the coffee will approach. For \(T(t) = 75\cdot(0.9)^{\frac{t}{5}}+20\), the horizontal asymptote is \(T = 20^{\circ}C\), indicating that as time progresses, the temperature of the coffee will approach 20°C.

Key concepts

Temperature Decay

In the context of Newton's Law of Cooling, temperature decay refers to how the temperature of a liquid decreases over time. This happens when a hot liquid, like coffee, is left to cool in a room. Initially the liquid is much hotter than the surrounding air, but over time the temperature of the liquid 'decays' and approaches the ambient temperature.
Newton's Law of Cooling can be described using the formula:\[T(t)=\left(T_{i}-T_{f}\right)(0.9)^{\frac{t}{5}}+T_{f},\]where:

• \(T_i\) – Initial temperature of the liquid
• \(T_f\) – Final temperature the liquid will approach
• \(t\) – Time in minutes

This formula shows that the temperature change is dependent on the time elapsed and the difference between the initial and final temperatures.
The cooling rate is faster when the initial temperature is much higher than the surrounding temperature, and slows down as the liquid gets closer to the ambient temperature.

Exponential Functions

Exponential functions are a key mathematical concept for understanding temperature decay. They are functions where a constant base is raised to a variable exponent. The general form an exponential function takes is:
\[f(x) = a \, b^x\]
In the case of Newton's Law of Cooling, the formula:
\[T(t) = \left(T_i - T_f\right) (0.9)^{\frac{t}{5}} + T_f\]
shows an exponential relationship between temperature and time. Here, the base of the exponent is 0.9 which indicates a decay factor. This factor is less than 1, meaning the temperature decreases exponentially over time.
When dealing with exponential decay, like in this problem, the quantity reduces by a consistent percentage over equal time intervals. For instance, in our scenario, the temperature reduces by a factor of 0.9 for every 5 minutes that passes.

Horizontal Asymptote

In the graph of the cooling function, there is a horizontal asymptote which represents a value that the temperature approaches but never quite reaches. The concept of a horizontal asymptote is essential in understanding the behavior of temperature decay over a long period.
In our problem, the horizontal asymptote is given by the final temperature \(T_f\), which in this case is 20°C. This is the temperature that the coffee will approach as time approaches infinity.
Conceptually, this means that after a very long time, the temperature of the coffee will be very close to 20°C but will not drop below it. The horizontal asymptote provides a bound to the cooling process, indicating that the cooling will slow down as the temperature gets closer to the ambient temperature.

Parameter Identification

When working with equations, it is important to identify the parameters correctly. In our temperature decay problem, the parameters are:

• \(a\) – The initial effect of temperature change: This is the difference between the initial and the final temperature, \(T_i - T_f\). For the coffee problem, \(a = 95 - 20 = 75\).
• \(b\) – The base of the exponent, which shows the cooling rate. Here, \(b = 0.9\).
• \(h\) – The horizontal shift, which is 0 in the given model since there's no horizontal shift in time \(t\).
• \(k\) – The vertical shift or the final temperature of the cooling process, \(T_f\), which is 20°C.

Correct parameter identification will help in accurately graphing the temperature decay and predicting temperatures at different times.
Understanding each parameter enables a deeper grasp of how each part of the function affects the overall behavior of the temperature change over time.