Question

If the surrounding air is kept at 25$^\circ$C and a body cools from 80$^\circ$C to 50$^\circ$C in 30 minutes, then temperature of the body after one hour will be

• A. 31.72$^\circ$C approximately
• B. 34.74$^\circ$C approximately
• C. 32.36$^\circ$C approximately
• D. 36.36$^\circ$C approximately

Hint:

Notice that the time value doubles from 30 minutes to 60 minutes. In exponential cooling processes, the ratio of the temperature difference relative to the surroundings stays constant over equal intervals of time: $\frac{\theta_1 - \theta_0}{\theta_0' - \theta_0} = \frac{\theta_2 - \theta_0}{\theta_1 - \theta_0}$. This means $\frac{50-25}{80-25} = \frac{\theta-25}{50-25} \implies \frac{25}{55} = \frac{\theta-25}{25}$, which simplifies directly to $\theta - 25 = \frac{625}{55} = \frac{125}{11} = 11.36 \implies \theta = 36.36^\circ\text{C}$. This constant ratio property avoids calculating the cooling constant $k$ entirely!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:

The problem presents a cooling process that follows Newton's Law of Cooling. We are given the constant surrounding ambient temperature ($\theta_0 = 25^\circ\text{C}$), an initial temperature drop from $80^\circ\text{C}$ to $50^\circ\text{C}$ over a 30-minute interval, and we need to determine the body's final temperature after a total elapsed time of 1 hour (60 minutes).

Step 2: Key Formula or Approach:
According to Newton's Law of Cooling, the integrated temperature relation over time is given by: $$\log_e(\theta - \theta_0) = -kt + C$$ Alternatively, for discrete time steps, we can use the linear approximation formula: $$\frac{\theta_{\text{initial}} - \theta_{\text{final}}}{t} = K \left( \frac{\theta_{\text{initial}} + \theta_{\text{final}}}{2} - \theta_0 \right)$$ Using the exact exponential/logarithmic solution ensures accurate results for longer time steps.

Step 3: Detailed Explanation:
Using the integrated formula: $\log_e(\theta - 25) = -kt + C$ 1. At $t = 0$ minutes, the initial temperature is $\theta = 80^\circ\text{C}$: $$\log_e(80 - 25) = -k(0) + C \implies C = \log_e(55)$$ 2. At $t = 30$ minutes, the temperature drops to $\theta = 50^\circ\text{C}$: $$\log_e(50 - 25) = -k(30) + \log_e(55)$$ $$\log_e(25) - \log_e(55) = -30k \implies \log_e\left(\frac{25}{55}\right) = -30k$$ $$\log_e\left(\frac{5}{11}\right) = -30k \implies 30k = \log_e\left(\frac{11}{5}\right)$$ 3. We need to find the temperature $\theta$ at $t = 60$ minutes: $$\log_e(\theta - 25) = -k(60) + \log_e(55)$$ $$\log_e(\theta - 25) = -2(30k) + \log_e(55)$$ Substitute the value of $30k$ from our second step: $$\log_e(\theta - 25) = -2\log_e\left(\frac{11}{5}\right) + \log_e(55) = \log_e\left(\frac{5}{11}\right)^2 + \log_e(55)$$ $$\log_e(\theta - 25) = \log_e\left(\frac{25}{121} \times 55\right)$$ Simplify the fraction inside the logarithm by dividing by 11: $$\log_e(\theta - 25) = \log_e\left(\frac{25 \times 5}{11}\right) = \log_e\left(\frac{125}{11}\right)$$ Remove the logarithms from both sides: $$\theta - 25 = \frac{125}{11} \approx 11.36$$ $$\theta = 25 + 11.36 = 36.36^\circ\text{C}$$

Step 4: Final Answer:
The temperature of the body after one hour will be approximately 36.36$^\circ$C, which corresponds to option (D).