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Chapter 9: Problem 14

Suppose you have just poured a cup of freshly brewed coffee with temperature \( 95^{\circ} \) in a room where the temperature is \( 20^{\circ}. \) (a) When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain. (b) Newtons Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newtons Law of Cooling for this particular situation. What is the initial condition? In view of your answer to part (a), do you think this differential equation is an appropriate model for cooling? (c) Make a rough sketch of the graph of the solution of the initial-value problem in part (b).

Short Answer

Expert verified
The coffee cools most quickly immediately after being poured; the rate of cooling slows over time. The differential equation is \( \frac{dT}{dt} = -k(T - 20) \) with initial condition \( T(0) = 95 \); it's suitable for the situation.

Step by step solution

01

Understanding Part (a)

The coffee will cool most quickly just after it is poured because the temperature difference between the coffee and the room is greatest at this point. As the coffee cools, this temperature difference decreases, leading to a slower rate of cooling over time. According to Newton's Law of Cooling, the rate of cooling is directly proportional to this temperature difference. Thus, the rate diminishes as the temperature difference decreases over time.
02

Formulating the Differential Equation for Part (b)

Newton's Law of Cooling can be expressed by the differential equation: \( \frac{dT}{dt} = -k(T - T_s) \), where \( T \) is the temperature of the coffee at time \( t \), \( T_s \) is the ambient temperature (\( 20^{\circ} \) in this case), and \( k \) is a positive constant. The initial condition is \( T(0) = 95^{\circ} \). This differential equation is appropriate for modeling the cooling process since the initial temperature difference between the coffee and the surroundings isn't extremely large, which is a condition for Newton's Law to be valid.
03

Sketching the Solution Graph for Part (c)

To sketch the solution to the initial-value problem, start at the point (0, 95) on the temperature vs. time graph. The graph should initially slope steeply downwards due to the large temperature difference. As time progresses, the slope (which represents the rate of cooling) decreases, and the curve approaches a horizontal asymptote at \( T_s = 20^{\circ} \), meaning the coffee approaches room temperature. The curve is an example of an exponential decay function, which indicates rapid initial cooling that slows over time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

differential equations
In mathematics, differential equations are equations that relate a function to its derivatives. They play a crucial role in modeling the behavior of natural phenomena, such as the cooling of coffee. For Newton's Law of Cooling, we use a specific type of differential equation. This equation, expressed as \(\frac{dT}{dt} = -k(T - T_s)\), tells us how the temperature \(T\) of an object changes over time \(t\). Here, \(T_s\) represents the surrounding temperature, and \(k\) is a proportionality constant that depends on the specific conditions of the system. The negative sign indicates that the temperature decreases, aligning with the idea of cooling.

Understanding differential equations involves recognizing that they inform us about the rate of change. In our coffee cooling problem, this equation helps us determine how quickly the coffee temperature drops towards the room temperature.
temperature difference
Temperature difference is a central concept in Newton鈥檚 Law of Cooling. It鈥檚 the driving force behind the rate of change mentioned in differential equations. Essentially, it measures how much warmer or cooler an object is compared to its surroundings.

When the temperature difference is large, as right after pouring the coffee, the cooling process is rapid. This is why initially, we observe the coffee to cool faster. Over time, as the temperature difference becomes smaller, the rate of cooling slows down proportionally. The cooling rate is always trying to minimize this temperature difference, working towards equalizing the temperature of the object with its environment.
exponential decay
Exponential decay is a mathematical concept that describes how quantities decrease rapidly at first and then slow over time. In the context of Newton鈥檚 Law of Cooling, the temperature of the coffee follows this pattern. As soon as it's poured, the temperature drops significantly due to the large temperature difference.

The equation \(T(t) = T_s + (T_0 - T_s) e^{-kt}\) models this behavior, where \(T_0\) is the initial temperature, \(T_s\) the surrounding temperature, and \(e\) is the base of natural logarithms. The term \(e^{-kt}\) represents the exponential decay factor that decreases over time, showing why the temperature change is significant at the start but diminishes later, asymptotically approaching the ambient temperature.
initial-value problem
An initial-value problem is a type of differential equation that requires a solution which meets pre-specified values at the outset. For the coffee cooling scenario, our initial condition is \(T(0) = 95^{\circ}\). This indicates that the initial temperature of the coffee is \(95^{\circ}\) at time \(t=0\).

Solving this initial-value problem means finding a function \(T(t)\) that not only satisfies the differential equation \(\frac{dT}{dt} = -k(T - T_s)\) but also aligns with this initial condition. Understanding this setup is crucial, as it determines the uniqueness of the solution and how accurately the function can model real-world cooling behavior, elegantly depicting the coffee's journey from hot to room temperature.

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Most popular questions from this chapter

In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product \( C: A + B \to C. \) The law of mass action states that the rate of reaction is proportional to the product of the concentrations of A and B: \( \frac {d[C]}{dt} = k [A][B] \) (See Examples 3.7.4.) Thus, if the initial concentrations are \( [A] = a \) moles/L and \( [B] = b \) moles/L and we write \( x = [C], \) then we have \( \frac {dx}{dt} = k(a - x)(b - x) \) (a) Assuming that \( a \neq b, \) find \( x \) as a function of \( t. \) Use the fact that the initial concentration of C is 0. (b) Find \( x(t) \) assuming that \( a = b. \) How does this expression for \( x(t) \) simplify if it is known that \( [C] = \frac {1}{2}a \) after 20 seconds?

An object with mass \( m \) is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If \( s(t) \) is the distance dropped after \( t \) seconds, then the speed is \( v = s'(t) \) and the acceleration is \( a = v'(t). \) If \( g \) is the acceleration due to gravity, them the downward force on the object is \( mg - cv, \) where \( c \) is a positive constant, and Newton's Second Law gives \( m \frac {dv}{dt} = mg - cv \) (a) Solve this as a linear equation to show that \( v = \frac {mg}{c} (1 - e^{-ct/m}) \) (b) What is the limiting velocity? (c) Find the distance the object has fallen after \( t \) seconds.

A model for tumor growth is given by the Gompertz equation \( \frac {dV}{dt} = a (\ln b - \ln V) V \) where \( a \) and \( b \) are positive constant and \( V \) is the volume of the tumor measured in \( mm^3. \) (a) Find a family of solution for tumor volume as a function of time. (b) Find the solution that has an initial tumor volume of \( V(0) = 1 mm^3. \)

The Pacific halibut fishery has been modeled by the differential equation \( \frac {dy}{dt} = ky (1 - \frac {y}{M}) \) where \( y(t) \) is the biomass (the total mass of the members of the population) in kilograms at time \( t \) (measured in years), the carrying capacity is estimated to be \( M = 8 \times 10^7 \) kg, and \( k = 0.71 \) per year. (a) If \( y(0) = 2 \times 10^7 \) kg, find the biomass a year later. (b) How long will it take for the biomass to reach \( 4 \times 10^7 \) kg?

A vat with 500 gallons of beer contains \( 4% \) alcohol (by volume). Beer with \( 6% \) alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?
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