Question

A student performs the following experiment using two identical cups of water. One cup is removed from a refrigerator at \(34^{\circ} \mathrm{F}\) and allowed to warm in its surroundings to room temperature \(\left(72^{\circ} \mathrm{F}\right)\). A second cup is simultaneously taken from room temperature surroundings and placed in the refrigerator to cool. The time at which each cup of water reached a temperature of \(53^{\circ} \mathrm{F}\) is recorded. Are the two recorded times the same or not? Explain.

Short answer

Answer: Yes, the two cups of water reach a temperature of 53°F at the same time.

Step-by-step solution

Understand Newton's Law of Cooling

Newton's Law of Cooling is given by the following differential equation: \(\frac{dT}{dt} = k(T - T_{ambient})\), where \(T\) is the temperature of the object, \(T_{ambient}\) is the ambient temperature, and \(k\) is a constant that depends on the properties of the object. At every point in time, the rate of cooling depends on the difference between the temperature of the object and that of the ambient temperature.

Make Observations for Both Cups

We need to analyze the initial and final temperatures in both scenarios: - Cup A (removed from the refrigerator): Initial temperature \(T_A = 34^{\circ}\mathrm{F}\), final temperature \(T_{A^*} = 53^{\circ}\mathrm{F}\) , ambient temperature \(T_{ambient} = 72^{\circ}\mathrm{F}\) - Cup B (put in the refrigerator): Initial temperature \(T_B = 72^{\circ}\mathrm{F}\), final temperature \(T_{B^*} = 53^{\circ}\mathrm{F}\), ambient temperature \(T_{ambient} = 34^{\circ}\mathrm{F}\)

Compare Temperature Differences for Both Cups

We can compare the temperature differences for both cups at various stages of the cooling: Initial Temperature Difference: - Cup A: \(T_{A} - T_{ambient} = -38^{\circ}\mathrm{F}\) - Cup B: \(T_{B} - T_{ambient} = 38^{\circ}\mathrm{F}\) Final Temperature Difference: - Cup A: \(T_{A^*} - T_{ambient} = -19^{\circ}\mathrm{F}\) - Cup B: \(T_{B^*} - T_{ambient} = 19^{\circ}\mathrm{F}\) Note that even though the initial and final temperature differences for Cup A and and Cup B are opposite in sign, their absolute values are equal.

Conclusion

According to Newton's Law of Cooling, the rate of cooling (or heating) depends on the difference between the initial and final temperatures. Since the absolute differences for both cups are equal, the rate of cooling (or heating) will be equal. Moreover, the cups of water are identical and the cooling rates depend only on the absolute difference between their initial and final temperatures. Consequently, the two cups of water reach a temperature of \(53^{\circ} \mathrm{F}\) at the same time. So, the recorded times for both cups to reach the target temperature of \(53^{\circ} \mathrm{F}\) are the same.

Key concepts

Differential Equations

Differential equations are mathematical tools that relate a function to its derivatives. They are essential in modeling the behavior of various physical systems, including the cooling and heating processes described by Newton's Law of Cooling. In the exercise, the differential equation \[\begin{equation}\frac{dT}{dt} = k(T - T_{ambient})\end{equation}\]
represents the rate at which an object's temperature changes over time. The equation encompasses the variable \[\begin{equation}T\end{equation}\]
for the object's temperature, a constant \[\begin{equation}k\end{equation}\]
related to the object's thermal properties, and the ambient temperature \[\begin{equation}T_{ambient}\end{equation}\]
. By solving this differential equation, one can predict how quickly an object will cool down or warm up to match the temperature of its surroundings. Students often find differential equations challenging, but understanding the relationship between the rate of change and the driving factors as shown in this equation is a fundamental skill in both mathematics and the applied sciences.

Ambient Temperature

Ambient temperature, often denoted as \[\begin{equation}T_{ambient}\end{equation}\]
, is a crucial concept when discussing Newton's Law of Cooling. It is the temperature of the surrounding environment into which the object is placed. If an object is warmer than the ambient temperature, it will lose heat; if it is cooler, it will gain heat. For example, in the provided exercise, the ambient temperature for Cup A is the room temperature of \[\begin{equation}72^{\textdegree} F\end{equation}\]
, while for Cup B, which is placed in the fridge, the ambient temperature is \[\begin{equation}34^{\textdegree} F\end{equation}\]
. These ambient temperatures play a foundational role in determining the rate at which the respective cup's temperature changes.

Rate of Cooling

The rate of cooling describes how quickly the temperature of an object decreases over time. Newton's Law tells us that this rate is not constant but varies depending on the temperature difference between the object and its ambient environment. The greater this difference, the faster the rate of cooling. This inverse relationship ensures that as an object's temperature approaches the ambient temperature, its rate of cooling decreases, leading to a slower change in temperature. The equation manifesting this relationship is a differential equation where \[\begin{equation}\frac{dT}{dt}\end{equation}\]
signifies the rate of temperature change. For both cups mentioned in the exercise, the cooling process (for Cup B) or the heating process (for Cup A) will proceed rapidly at first and then slow down as the object's temperature gets closer to the ambient level.

Temperature Differences

Temperature differences are the driving force of thermal change according to Newton's Law of Cooling. In the context of the exercise, we observe this by comparing the initial and final temperature differences of the cups. Cup A, starting with \[\begin{equation}34^{\textdegree} F\end{equation}\]
warms up towards the ambient temperature, and Cup B cools down from \[\begin{equation}72^{\textdegree} F\end{equation}\]
toward the colder fridge environment. The absolute values of these temperature differences are equal but opposite in sign. This means that, though Cup A and Cup B are in thermally opposite processes, they both reach the midpoint temperature of \[\begin{equation}53^{\textdegree} F\end{equation}\]
at the same time. Students should note that it is the magnitude of temperature differences, not their algebraic sign, that dictates the rate of thermal change.