Question

If a body cools from $80^{\circ}\text{C}$ to $50^{\circ}\text{C}$ in the room temperature of $25^{\circ}\text{C}$ in 30 minutes, then the temperature of the body after 1 hour is

• A. $31.36^{\circ}\text{C}$
• B. $32.25^{\circ}\text{C}$
• C. $36.36^{\circ}\text{C}$
• D. $33.25^{\circ}\text{C}$

Hint:

For Newton's Law of Cooling, always remember the formula $\theta(t) = \theta_s + Ce^{-kt}$. Use initial conditions to find $C$ and then another data point to find $k$. If time is doubled, the cooling factor $e^{-kt}$ becomes squared.

Answer 1

The Correct Option is C

Step 1: State Newton's Law of Cooling. Newton's Law of Cooling states that the rate of change of temperature of a body is proportional to the difference between its own temperature and the ambient temperature (room temperature). Let $\theta(t)$ be the temperature of the body at time $t$, and $\theta_s$ be the surrounding temperature. \[\frac{d\theta}{dt} = -k(\theta - \theta_s)\] Here, the room temperature $\theta_s = 25^{\circ}\text{C}$. So, the differential equation is: \[\frac{d\theta}{dt} = -k(\theta - 25)\]
Step 2: Integrate the differential equation. Separate variables and integrate: \[\frac{d\theta}{\theta - 25} = -k\,dt\] Integrating both sides: \[\int \frac{d\theta}{\theta - 25} = \int -k\,dt\] \[\ln|\theta - 25| = -kt + C_1\] Exponentiate both sides: \[|\theta - 25| = e^{-kt + C_1}\] \[\theta - 25 = \pm e^{C_1}e^{-kt}\] Let $C = \pm e^{C_1}$. Then: \[\theta(t) = 25 + Ce^{-kt}\]
Step 3: Use the initial condition to find $C$. At $t = 0$ minutes, the initial temperature is $\theta(0) = 80^{\circ}\text{C}$. \[80 = 25 + Ce^{-k(0)}\] \[80 = 25 + C\] \[C = 80 - 25 = 55\] So the equation becomes: \[\theta(t) = 25 + 55e^{-kt}\]
Step 4: Use the given condition ($t=30$ minutes, $\theta=50^{\circ}\text{C}$) to find the cooling constant $k$. At $t = 30$ minutes, $\theta(30) = 50^{\circ}\text{C}$. \[50 = 25 + 55e^{-k(30)}\] \[25 = 55e^{-30k}\] \[e^{-30k} = \frac{25}{55} = \frac{5}{11}\] Take the natural logarithm of both sides: \[-30k = \ln\left(\frac{5}{11}\right)\] \[k = -\frac{1}{30}\ln\left(\frac{5}{11}\right) = \frac{1}{30}\ln\left(\frac{11}{5}\right)\]
Step 5: Calculate the temperature after 1 hour (60 minutes). We need to find $\theta(60)$. From Step 4, we have $e^{-30k} = \frac{5}{11}$. We can write $e^{-60k}$ as $(e^{-30k})^2$. \[e^{-60k} = \left(\frac{5}{11}\right)^2 = \frac{25}{121}\] Now substitute this into the temperature equation: \[\theta(60) = 25 + 55e^{-60k}\] \[\theta(60) = 25 + 55\left(\frac{25}{121}\right)\] \[\theta(60) = 25 + \frac{55 \times 25}{121}\] Simplify the fraction: $55 = 5 \times 11$ and $121 = 11 \times 11$. \[\theta(60) = 25 + \frac{5 \times 11 \times 25}{11 \times 11}\] \[\theta(60) = 25 + \frac{5 \times 25}{11}\] \[\theta(60) = 25 + \frac{125}{11}\] Convert $\frac{125}{11}$ to a decimal: \[\frac{125}{11} \approx 11.3636...\] So, \[\theta(60) \approx 25 + 11.3636\] \[\theta(60) \approx 36.36^{\circ}\text{C}\]