Differential Equations Quiz: Newtons Law Of Cooling Heating

What this quiz covers

This quiz focuses on Newtons Law Of Cooling Heating, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $t_1 < t_2$ (correct answer)
2. $t_1 > t_2$
3. $t_1 = t_2$
4. The relationship depends on the ambient temperature.

Explanation: 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 $50°C$.

Question 2

1. $40.00°C$
2. $26.88°C$
3. $48.75°C$
4. $28.75°C$ (correct answer)

Explanation: 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 $k$: $T'(30) = 5 + 95e^{(-\ln(2)/15) \cdot 30} = 5 + 95e^{-2\ln(2)} = 5 + 95e^{\ln(2^{-2})} = 5 + 95(1/4) = 5 + 23.75 = 28.75°C$.

Question 3

1. $5/3$
2. $5/6$
3. $1$ (correct answer)
4. $0$

Explanation: 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 $\lim_{t \to \infty} \frac{T_A(t)}{T_B(t)} = \frac{\lim_{t \to \infty} T_A(t)}{\lim_{t \to \infty} T_B(t)} = \frac{20}{20} = 1$.

Question 4

1. $13.5°C$
2. $45.1°C$
3. $38.5°C$ (correct answer)
4. $55.0°C$

Explanation: 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.

Question 5

1. $10 \ln(3/5)$
2. $6 \ln(3/5)$
3. $(50/3) \ln(5/3)$
4. $(50/3) \ln(3/5)$ (correct answer)

Explanation: 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: $(dT/dt)|_{t=0} = (\frac{1}{5}\ln(3/5)) (250/3) = \frac{50}{3}\ln(3/5)$.

Question 6

1. The graph of the soup's temperature $T(t)$ versus time $t$ has a constant, non-zero concavity.
2. The average rate of change of $T(t)$ over any time interval is equal to the instantaneous rate of change at the interval's midpoint.
3. 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. (correct answer)
4. If the initial temperature of the soup is doubled, the time it takes to cool to a specific temperature $T_f$ (where $A < T_f < T(0)$) is also doubled.

Explanation: 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.

Question 7

1. $5 \ln(4)$ minutes
2. $10 \ln(2)$ minutes (correct answer)
3. $5 \ln(5/3)$ minutes
4. $(1/16) \ln(4)$ minutes

Explanation: 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 $t = 5\ln(4) = 5\ln(2^2) = 10\ln(2)$.

Question 8

1. $k_2 = 2k_1$
2. $k_2 = k_1/2$ (correct answer)
3. $k_2 = 4k_1$
4. $k_2 = k_1/4$

Explanation: 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.

Question 9

1. $\frac{dT}{dt} = -k(T - 80), \quad T(0) = 20$
2. $\frac{dT}{dt} = k(T - 20), \quad T(0) = 80$
3. $\frac{dT}{dt} = -k(T - 20), \quad T(0) = 60$
4. $\frac{dT}{dt} = -k(T - 20), \quad T(0) = 80$ (correct answer)

Explanation: 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 $T(0) = 80$.

Question 10

1. $10$ minutes
2. $10 \frac{\ln(2)}{\ln(3/2)}$ minutes (correct answer)
3. $10 \frac{\ln(3)}{\ln(3/2)}$ minutes
4. $10 \frac{\ln(1/2)}{\ln(2/3)}$ minutes

Explanation: 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, $\Delta t = \frac{\ln(1/2)}{k} = \frac{\ln(1/2)}{(1/10)\ln(2/3)} = 10 \frac{-\ln(2)}{\ln(2)-\ln(3)} = 10 \frac{\ln(2)}{\ln(3)-\ln(2)} = 10 \frac{\ln(2)}{\ln(3/2)}$.

Question 11

1. $15°C$
2. $20°C$ (correct answer)
3. $25°C$
4. $30°C$

Explanation: 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 $A=20$.

Question 12

1. $\frac{\ln(7/5)}{\ln(9/7)}$ (correct answer)
2. $\frac{\ln(9/7)}{\ln(7/5)}$
3. $7/9$
4. $5/9$

Explanation: 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 $\frac{-\ln(7/5)}{-\ln(9/7)} = \frac{\ln(7/5)}{\ln(9/7)}$.

Question 13

1. $21.25°C$ (correct answer)
2. $22.75°C$
3. $20.50°C$
4. $23.44°C$

Explanation: 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: $T(5) = 25 - 20e^{-5k} = 25 - 20e^{-5(\ln(2)/2)} = 25 - 20e^{-2.5\ln(2)} = 25 - 20(2^{-2.5}) = 25 - 20(\frac{1}{2^{2.5}}) = 25 - 20(\frac{1}{4\sqrt{2}}) = 25 - \frac{20}{4\sqrt{2}} = 25 - \frac{5}{\sqrt{2}} \approx 25 - 3.75 = 21.25°C$.

Question 14

1. $95°C$
2. $100°C$ (correct answer)
3. $90°C$
4. $105°C$

Explanation: 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 $T_o = 100°C$.

Question 15

1. Approximately 4.8 minutes (correct answer)
2. Approximately 6.2 minutes
3. Approximately 5.5 minutes
4. Approximately 3.9 minutes

Explanation: 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.

Question 16

1. Approximately 58.4%
2. Approximately 77.7% (correct answer)
3. Approximately 63.2%
4. Approximately 69.4%

Explanation: 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 $\frac{46.61}{60} \times 100\% = 77.7\%$.