Question

A bowl filled with very hot soup cools from $98^{\circ} C$ to $86^{\circ} C$ in 2 minutes when the room temperature is $22^{\circ} C$. How long it will take to cool from $75^{\circ} C$ to $69^{\circ} C$ ?

• A. 2 minutes
• B. 1 minute
• C. $1.4$ minutes
• D. $0.5$ minute

Answer 1

The Correct Option is C

The problem involves cooling of the soup, which can be analyzed using Newton's Law of Cooling. This law states:

\(T(t) = T_{\text{ambient}} + (T_0 - T_{\text{ambient}}) e^{-kt}\)

Where:

• \(T(t)\) is the temperature of the object at time \(t\),
• \(T_{\text{ambient}}\) is the ambient room temperature,
• \(T_0\) is the initial temperature of the object,
• \(k\) is the cooling constant.

Let's apply this law to both situations given in the problem:

1. First Cooling Event: The temperature drops from \(98^{\circ}C\) to \(86^{\circ}C\) in 2 minutes.
   • Initial temperature, \(T_0 = 98^{\circ}C\)
   • Final temperature after 2 minutes, \(T(2) = 86^{\circ}C\)
   • Room temperature, \(T_{\text{ambient}} = 22^{\circ}C\)
2. Using Newton's Law of Cooling, we can write: \(86 = 22 + (98 - 22)e^{-2k}\)
3. Solving for \(e^{-2k}\), we have: \(64 = 76e^{-2k}\)
4. This gives us: \(e^{-2k} = \frac{64}{76} = \frac{16}{19}\)
5. Second Cooling Event: The temperature drops from \(75^{\circ}C\) to \(69^{\circ}C\).
   • Initial temperature, \(T_0 = 75^{\circ}C\)
   • Final temperature, \(T(t)= 69^{\circ}C\)
6. Apply Newton's Law of Cooling: \(69 = 22 + (75 - 22)e^{-kt}\)
7. Solving for \(e^{-kt}\), we have: \(47 = 53e^{-kt}\)
8. This gives us: \(e^{-kt} = \frac{47}{53}\)
9. Now, relate both time frames: \((e^{-kt})^2 = e^{-2k}\)
10. Therefore, \(\left(\frac{47}{53}\right)^2 = \frac{16}{19}\)
11. Solving for \(t\), we find: \(t = 2 \times \frac{\ln\left(\frac{47}{53}\right)}{\ln\left(\frac{16}{19}\right)}\)
12. Calculating gives approximately: \(t \approx 1.4 \text{ minutes}\)

Therefore, the time it takes for the soup to cool from \(75^{\circ}C\) to \(69^{\circ}C\) is approximately 1.4 minutes.