Newton's Law of Cooling/Heating - Differential Equations

1. A pie at $180°C$ is removed from an oven and left in a room at $20°C$. After 15 minutes, its temperature is $100°C$. At this moment, the pie is moved to a refrigerator at $5°C$. What is the temperature of the pie 30 minutes after it was placed in the refrigerator?
   Answer: $28.75°C$ First, find the cooling constant $k$ in the room. $T(t) = A_1 + (T_0 - A_1)e^{kt}$. With $A_1=20$, $T_0=180$, and $T(15)=100$, we have $100 = 20 + (180-20)e^{15k}$. This gives $80 = 160e^{15k}$, so $e^{15k} = 1/2$, and $k = \ln(1/2)/15 = -\ln(2)/15$. Now, for the second phase in the refrigerator, we have a new problem with $A_2=5$ and a new initial temperature $T'_0=100$. We want to find the temperature after 30 minutes, i.e., $T'(30)$. The solution is $T'(t') = A_2 + (T'_0-A_2)e^{kt'} = 5 + (100-5)e^{k(30)}$. Substituting $k$: $T'(30) = 5 + 95e^{(-\ln(2)/15) \cdot 30} = 5 + 95e^{-2\ln(2)} = 5 + 95e^{\ln(2^{-2})} = 5 + 95(1/4) = 5 + 23.75 = 28.75°C$.
2. The temperature $T(t)$ of a cooling object is given by the function $T(t) = 20 + 60e^{-kt}$ for some positive constant $k$. Which of the following initial value problems accurately models the object's temperature?
   Answer: $\frac{dT}{dt} = -k(T - 20), \quad T(0) = 80$ The general solution to Newton's Law of Cooling, $\frac{dT}{dt} = c(T-A)$, is $T(t) = A + (T_0-A)e^{ct}$. From the given function $T(t) = 20 + 60e^{-kt}$, we can identify the parameters. As $t \to \infty$, $e^{-kt} \to 0$, so $T(t) \to 20$. This means the ambient temperature is $A=20$. At $t=0$, the initial temperature is $T(0) = 20 + 60e^0 = 20 + 60 = 80$. Thus, $T_0=80$. Comparing the exponential terms $e^{ct}$ and $e^{-kt}$, we see that the rate constant in the differential equation is $c = -k$. Therefore, the differential equation is $\frac{dT}{dt} = -k(T - 20)$. The initial value problem is $\frac{dT}{dt} = -k(T - 20)$ with the initial condition $T(0) = 80$.
3. A container of hot soup is cooling in a room with a constant ambient temperature $A$. The temperature difference between the soup and the room is given by the function $D(t) = T(t) - A$. Which of the following statements about the cooling process is necessarily true?
   Answer: The time required for $D(t)$ to reduce to one-quarter of its value is double the time required for it to reduce to one-half of its value. When analyzing cooling processes, you're dealing with Newton's Law of Cooling, which states that the rate of temperature change is proportional to the temperature difference: $\frac{dD}{dt} = -kD$, where $k > 0$. This gives us the exponential decay solution $D(t) = D_0 e^{-kt}$. Option C is correct because exponential decay has a specific half-life property. If $D(t)$ reduces to half its value at time $t_{1/2}$, then $D_0/2 = D_0 e^{-kt_{1/2}}$, so $t_{1/2} = \frac{\ln(2)}{k}$. For it to reduce to one-quarter, we need $D_0/4 = D_0 e^{-kt_{1/4}}$, giving $t_{1/4} = \frac{\ln(4)}{k} = \frac{2\ln(2)}{k} = 2t_{1/2}$. The quarter-time is exactly double the half-time. Option A is wrong because while $T(t) = A + D_0 e^{-kt}$ has constant concavity (it's always concave up since the second derivative $kD_0 e^{-kt} > 0$), this isn't necessarily true for all cooling models. Option B is incorrect because exponential functions don't satisfy the Mean Value Theorem in this special way. The average rate over an interval generally differs from the instantaneous rate at the midpoint. Option D fails because doubling the initial temperature changes $D_0$ but doesn't create a linear scaling relationship for reaching a specific final temperature. The exponential nature means the relationship isn't proportional. Remember: exponential decay processes have predictable timing ratios. The time to reach any fraction is always a constant multiple of the half-life, regardless of starting conditions.
4. A liquid initially at $40°C$ is heated in an oven. The oven temperature is unknown, but the liquid reaches $60°C$ after 5 minutes and $75°C$ after 10 minutes. Assuming Newton's law of heating applies, what is the oven temperature?
   Answer: $100°C$ Let $T_o$ be the oven temperature. Using $T(t) = T_o + (40-T_o)e^{-kt}$. At $t=5$: $60 = T_o + (40-T_o)e^{-5k}$, so $60-T_o = (40-T_o)e^{-5k}$. At $t=10$: $75 = T_o + (40-T_o)e^{-10k}$, so $75-T_o = (40-T_o)e^{-10k}$. Since $e^{-10k} = (e^{-5k})^2$: $\frac{75-T_o}{40-T_o} = (\frac{60-T_o}{40-T_o})^2$. Let $x = \frac{60-T_o}{40-T_o}$, then $\frac{75-T_o}{40-T_o} = x^2$. From the first equation: $60-T_o = x(40-T_o)$, so $x = \frac{60-T_o}{40-T_o}$. From the second: $75-T_o = x^2(40-T_o)$. Substituting: $75-T_o = (\frac{60-T_o}{40-T_o})^2(40-T_o) = \frac{(60-T_o)^2}{40-T_o}$. Cross-multiplying: $(75-T_o)(40-T_o) = (60-T_o)^2$. Expanding: $3000 - 75T_o - 40T_o + T_o^2 = 3600 - 120T_o + T_o^2$. Simplifying: $3000 - 115T_o = 3600 - 120T_o$, so $5T_o = 600$, giving $T_o = 100°C$.
5. An object follows Newton's law of cooling with a time constant of $\tau = 12$ minutes (where $k = \frac{1}{\tau}$). If the object starts at $80°C$ in a $20°C$ environment, what percentage of the initial temperature difference has been lost after 18 minutes?
   Answer: Approximately 77.7% The initial temperature difference is $80 - 20 = 60°C$. Using $T(t) = 20 + 60e^{-t/\tau}$ with $\tau = 12$ minutes. After 18 minutes: $T(18) = 20 + 60e^{-18/12} = 20 + 60e^{-1.5} = 20 + 60e^{-3/2}$. Since $e^{-3/2} \approx 0.2231$, we get $T(18) = 20 + 60(0.2231) = 20 + 13.39 = 33.39°C$. The remaining temperature difference is $33.39 - 20 = 13.39°C$. The lost temperature difference is $60 - 13.39 = 46.61°C$. The percentage lost is $\frac{46.61}{60} \times 100\% = 77.7\%$.
6. A cup of coffee at $90°C$ is placed in a room at $20°C$. It takes $t_1$ minutes for the coffee to cool to $70°C$. A second, identical cup of coffee, also at $90°C$, is placed in a freezer at $0°C$. It takes $t_2$ minutes for this second cup to cool to $70°C$. Assuming the cooling constant $k$ is the same in both situations, what is the ratio $t_1/t_2$?
   Answer: $\frac{\ln(7/5)}{\ln(9/7)}$ For the first cup (in the room): $A_1=20, T_0=90, T(t_1)=70$. The equation is $70-20 = (90-20)e^{kt_1}$, which simplifies to $50 = 70e^{kt_1}$, so $e^{kt_1} = 5/7$. Taking logs, $kt_1 = \ln(5/7)$. For the second cup (in the freezer): $A_2=0, T_0=90, T(t_2)=70$. The equation is $70-0 = (90-0)e^{kt_2}$, which simplifies to $70 = 90e^{kt_2}$, so $e^{kt_2} = 7/9$. Taking logs, $kt_2 = \ln(7/9)$. To find the ratio $t_1/t_2$, we divide the two results: $\frac{t_1}{t_2} = \frac{\ln(5/7)/k}{\ln(7/9)/k} = \frac{\ln(5/7)}{\ln(7/9)}$. Using logarithm properties, this is equivalent to $\frac{-\ln(7/5)}{-\ln(9/7)} = \frac{\ln(7/5)}{\ln(9/7)}$.
7. A substance cools from $80°C$ to $60°C$ in 10 minutes in a room with an ambient temperature of $20°C$. How much additional time will it take for the substance to cool to $40°C$?
   Answer: $10 \frac{\ln(2)}{\ln(3/2)}$ minutes Let $T(t)$ be the temperature and $A=20$. The solution is $T(t) - A = (T_0 - A)e^{kt}$. First interval: $T_0 = 80$. At $t=10$, $T(10)=60$. So, $60-20 = (80-20)e^{10k}$, which gives $40 = 60e^{10k}$, so $e^{10k} = 2/3$. Let $\Delta t$ be the additional time to cool from $60°C$ to $40°C$. At time $10+\Delta t$, the temperature is $40°C$. The initial temperature for this phase can be considered $60°C$ at time $t'=0$. Then $40-20 = (60-20)e^{k\Delta t}$, which gives $20 = 40e^{k\Delta t}$, so $e^{k\Delta t} = 1/2$. From the first part, $10k = \ln(2/3)$, so $k = \frac{1}{10}\ln(2/3)$. From the second part, $k\Delta t = \ln(1/2)$. So, $\Delta t = \frac{\ln(1/2)}{k} = \frac{\ln(1/2)}{(1/10)\ln(2/3)} = 10 \frac{-\ln(2)}{\ln(2)-\ln(3)} = 10 \frac{\ln(2)}{\ln(3)-\ln(2)} = 10 \frac{\ln(2)}{\ln(3/2)}$.
8. Object A, initially at $100°C$, and Object B, initially at $120°C$, are placed in a large room with a constant ambient temperature of $20°C$. The cooling constant for A is $k_A = -0.05 \text{ min}^{-1}$ and for B is $k_B = -0.03 \text{ min}^{-1}$. Let $T_A(t)$ and $T_B(t)$ be their respective temperatures at time $t$. What is the value of the limit $\lim_{t \to \infty} \frac{T_A(t)}{T_B(t)}$?
   Answer: $1$ According to Newton's Law of Cooling, as time $t$ approaches infinity, the temperature of any object will approach the ambient temperature of its surroundings. In this case, both Object A and Object B are in a room with an ambient temperature of $20°C$. Therefore, $\lim_{t \to \infty} T_A(t) = 20$ and $\lim_{t \to \infty} T_B(t) = 20$. The initial temperatures and cooling constants affect the rate of approach to the ambient temperature, but not the final limiting value. The limit of the ratio is then $\lim_{t \to \infty} \frac{T_A(t)}{T_B(t)} = \frac{\lim_{t \to \infty} T_A(t)}{\lim_{t \to \infty} T_B(t)} = \frac{20}{20} = 1$.
9. A thermometer reading $5°C$ is brought into a room where the temperature is $25°C$. After 2 minutes, the thermometer reads $15°C$. According to Newton's law of heating, what will the thermometer read after 5 minutes total?
   Answer: $21.25°C$ Using Newton's law: $T(t) = 25 + (5-25)e^{-kt} = 25 - 20e^{-kt}$. After 2 minutes: $15 = 25 - 20e^{-2k}$, so $10 = 20e^{-2k}$, giving $e^{-2k} = 0.5$. Therefore $k = \frac{\ln(2)}{2} \approx 0.3466$. After 5 minutes: $T(5) = 25 - 20e^{-5k} = 25 - 20e^{-5(\ln(2)/2)} = 25 - 20e^{-2.5\ln(2)} = 25 - 20(2^{-2.5}) = 25 - 20(\frac{1}{2^{2.5}}) = 25 - 20(\frac{1}{4\sqrt{2}}) = 25 - \frac{20}{4\sqrt{2}} = 25 - \frac{5}{\sqrt{2}} \approx 25 - 3.75 = 21.25°C$.
10. A temperature probe initially reads $0°C$ when moved from a freezer to a laboratory at $22°C$. Due to the probe's thermal mass, it follows Newton's law of heating with a time constant of 90 seconds. If an accurate reading requires the probe to be within $1°C$ of the true temperature, how long must you wait before taking a measurement?
    Answer: Approximately 4.8 minutes The probe follows $T(t) = 22(1 - e^{-t/90})$ where $t$ is in seconds. For an accurate reading, we need $T(t) \geq 21°C$ (within 1°C of 22°C). So $22(1 - e^{-t/90}) \geq 21$, which gives $1 - e^{-t/90} \geq \frac{21}{22}$, or $e^{-t/90} \leq \frac{1}{22}$. Taking natural log: $-\frac{t}{90} \leq \ln(\frac{1}{22}) = -\ln(22)$, so $t \geq 90\ln(22) \approx 90 \times 3.091 = 278.2$ seconds $\approx 4.6$ minutes. The closest answer is 4.8 minutes.
11. A block of metal is placed in a room with an ambient temperature of $20°C$. After 5 minutes, its temperature is $70°C$, and after 10 minutes, its temperature is $50°C$. What was the initial rate of change of the temperature, $(dT/dt)|_{t=0}$, in $°C$ per minute?
    Answer: $(50/3) \ln(3/5)$ Let $T_0$ be the initial temperature. We have $T(t) - 20 = (T_0 - 20)e^{kt}$. At $t=5$: $70 - 20 = (T_0 - 20)e^{5k} \implies 50 = (T_0 - 20)e^{5k}$. At $t=10$: $50 - 20 = (T_0 - 20)e^{10k} \implies 30 = (T_0 - 20)e^{10k}$. Dividing the second equation by the first gives $30/50 = e^{5k}$, so $e^{5k} = 3/5$. This implies $k = \frac{1}{5}\ln(3/5)$. Substitute $e^{5k}=3/5$ into the first equation: $50 = (T_0-20)(3/5)$, which gives $T_0-20 = 250/3$. The initial temperature was $T_0 = 20 + 250/3 = 310/3$. The rate of change is $dT/dt = k(T-A)$. The initial rate is $(dT/dt)|_{t=0} = k(T_0-A) = k(T_0-20)$. Substituting the values we found: $(dT/dt)|_{t=0} = (\frac{1}{5}\ln(3/5)) (250/3) = \frac{50}{3}\ln(3/5)$.
12. An object cools in a room with a constant ambient temperature. Let $t_1$ be the time it takes for the object to cool from $90°C$ to $70°C$. Let $t_2$ be the time it takes for the object to cool further from $70°C$ to $50°C$. Which of the following statements correctly relates $t_1$ and $t_2$?
    Answer: $t_1 < t_2$ According to Newton's Law of Cooling, the rate of cooling is proportional to the temperature difference between the object and its surroundings. In the first interval, the temperature drops from $90°C$ to $70°C$. In the second interval, it drops from $70°C$ to $50°C$. The average temperature during the first interval is higher than the average temperature during the second interval. Therefore, the temperature difference with the ambient surroundings is greater during the first interval, leading to a faster rate of cooling. A faster rate of cooling over the same temperature drop ($20°C$) means it takes less time. Thus, $t_1 < t_2$. This is true regardless of the specific ambient temperature, as long as it is below $50°C$.
13. A hot object is placed in a room to cool. Its temperature is measured to be $100°C$ initially, $80°C$ after 10 minutes, and $65°C$ after 20 minutes. Assuming the object's temperature follows Newton's Law of Cooling, what is the ambient temperature of the room?
    Answer: $20°C$ Let $T(t)$ be the temperature and $A$ be the ambient temperature. The solution to Newton's Law of Cooling is $T(t) = A + (T_0 - A)e^{kt}$. We are given $T(0) = 100$, $T(10) = 80$, and $T(20) = 65$. Let $u(t) = T(t) - A$. Then $u(t) = u(0)e^{kt}$. We have $u(0) = 100-A$, $u(10) = 80-A$, and $u(20) = 65-A$. Since the time intervals are equal, the ratio of consecutive temperature differences must be constant: $u(10)/u(0) = u(20)/u(10) = e^{10k}$. Thus, $(80-A)/(100-A) = (65-A)/(80-A)$. Cross-multiplying gives $(80-A)^2 = (100-A)(65-A)$. Expanding both sides yields $6400 - 160A + A^2 = 6500 - 165A + A^2$. Simplifying this equation gives $5A = 100$, so $A=20$.
14. A metal object is heated to $150°F$ and placed in a room with a constant ambient temperature of $70°F$. Initially, the object is cooling at a rate of $16°F$ per minute. How long will it take for the object's temperature to reach $90°F$?
    Answer: $10 \ln(2)$ minutes Newton's Law of Cooling is given by the differential equation $dT/dt = k(T-A)$. We are given $A=70°F$, $T(0)=150°F$, and $(dT/dt)|_{t=0} = -16°F/min$. At $t=0$, we have $-16 = k(150 - 70) = 80k$, which implies $k = -16/80 = -1/5$. The solution is $T(t) = A + (T_0-A)e^{kt} = 70 + (150-70)e^{-t/5} = 70 + 80e^{-t/5}$. We want to find $t$ when $T(t)=90$. So, $90 = 70 + 80e^{-t/5}$, which simplifies to $20 = 80e^{-t/5}$, or $1/4 = e^{-t/5}$. Taking the natural logarithm of both sides gives $\ln(1/4) = -t/5$, so $-\ln(4) = -t/5$. This gives $t = 5\ln(4) = 5\ln(2^2) = 10\ln(2)$.
15. Two spheres, Sphere 1 and Sphere 2, are made of the same material and are initially at the same temperature. They are placed in the same room to cool. Sphere 1 has radius $r$ and Sphere 2 has radius $2r$. For an object, the cooling constant $k$ in Newton's Law of Cooling is proportional to its surface-area-to-volume ratio. If $k_1$ and $k_2$ are the cooling constants for Sphere 1 and Sphere 2, respectively, what is the relationship between them?
    Answer: $k_2 = k_1/2$ The surface area of a sphere is $S = 4\pi r^2$ and its volume is $V = \frac{4}{3}\pi r^3$. The surface-area-to-volume ratio is $S/V = (4\pi r^2) / (\frac{4}{3}\pi r^3) = 3/r$. The cooling constant $k$ is proportional to this ratio, so $k = c(3/r)$ for some constant of proportionality $c$. For Sphere 1, $k_1 = c(3/r)$. For Sphere 2, the radius is $2r$, so $k_2 = c(3/(2r))$. Comparing the two expressions, we see that $k_2 = \frac{1}{2} \left( c(3/r) \right) = \frac{1}{2}k_1$. This means the larger sphere cools more slowly.
16. A researcher measures the temperature $T(t)$ of a sample cooling in a lab maintained at $25°C$. A plot of $\ln(T(t) - 25)$ versus time $t$ (in minutes) is found to be a straight line with the equation $y = -0.04t + 3.0$. What is the approximate temperature of the sample at $t=10$ minutes?
    Answer: $38.5°C$ When you see a problem involving cooling with a linear relationship between $\ln(T - T_{\text{ambient}})$ and time, you're dealing with Newton's Law of Cooling. The key insight is recognizing how to work backwards from the linearized form to find the actual temperature. Given that $\ln(T(t) - 25) = -0.04t + 3.0$, you can find $T(10)$ by substituting $t = 10$: $\ln(T(10) - 25) = -0.04(10) + 3.0 = -0.4 + 3.0 = 2.6$ To solve for $T(10)$, take the exponential of both sides: $T(10) - 25 = e^{2.6} \approx 13.46$ Therefore: $T(10) = 13.46 + 25 = 38.46°C \approx 38.5°C$ Looking at the wrong answers: Choice A (13.5°C) represents the common error of forgetting to add back the ambient temperature—this is just $e^{2.6}$ without adding 25. Choice B (45.1°C) likely comes from calculation errors in the exponential. Choice D (55.0°C) might result from sign errors or mishandling the linear equation. The critical trap here is remembering that the linearized form is $\ln(T - T_{\text{ambient}})$, not $\ln(T)$. After finding the exponential, you must add back the ambient temperature to get the actual temperature. Always check that your final answer makes physical sense—the sample should be warmer than the room temperature but cooling down from its initial state.

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