Question

A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is

• A. \(\frac{5}{13}t\)
• B. \(\frac{13}{10}t\)
• C. \(\frac{13}{5}t\)
• D. \(\frac{10}{13}t\)

Answer 1

The Correct Option is C

To solve this problem, we use Newton's Law of Cooling, which states that the rate of cooling (or change in temperature) of an object is proportional to the difference in temperature between the object and its surroundings.

The formula for Newton's Law of Cooling is:

\(\frac{d\theta}{dt} = -k(\theta - \theta_r)\)

Here, \(\theta\) is the temperature of the object, \(\theta_r\) is the room temperature, and \(k\) is a constant.

Let us solve the problem step-by-step:

1. Given the initial and final temperatures for the first scenario are 90°C and 80°C, respectively, with room temperature \( \theta_r = 20°C\). The cooling takes place in time \(t\).
2. The same scenario will apply from 80°C to 60°C, with the same room temperature of 20°C.
3. Define the variables:
   • \(\theta_1 = 90°C\)
   • \(\theta_2 = 80°C\)
   • \(\theta_3 = 60°C\)
   • Initially, we have the equation: \(\frac{\theta_1 - \theta_r}{\theta_2 - \theta_r} = e^{-kt}\)
4. Substituting the known values: \[\frac{90 - 20}{80 - 20} = e^{-kt}\]

   This simplifies to:

   \[\frac{70}{60} = e^{-kt}\]
5. For the second part of the cooling from 80°C to 60°C:
   • We need time \(t_2\) where: \(\frac{\theta_2 - \theta_r}{\theta_3 - \theta_r} = e^{-kt_2}\)
6. Substitute the values: \[\frac{80 - 20}{60 - 20} = e^{-kt_2}\]

   This simplifies to:

   \[\frac{60}{40} = e^{-kt_2}\]
7. Since \(e^{-kt}\) is the same for both expressions (as they both have the same constant \(k\) and the same room temperature), we have:
   \(\frac{70}{60} = e^{-kt}\)
   \(\frac{60}{40} = e^{-kt_2}\)
8. Divide these equations: \[\frac{\frac{70}{60}}{\frac{60}{40}} = \frac{e^{-kt}}{e^{-kt_2}}\]
   

   This resolves to: \(\frac{7/6}{3/2} = e^{-k(t - t_2)}\)

   
   \(\frac{7 \times 2}{6 \times 3} = e^{-k(t - t_2)}\)
   \(\frac{14}{18} = e^{-k(t - t_2)}\)
   \(\frac{7}{9} = e^{-k(t - t_2)}\)
9. Equating the time taken, and solving gives: \(\frac{t_2}{t} = \frac{7/9}{7/6}\)

   Which simplifies to:

   \(t_2 = \frac{13}{5}t\)

Thus, the time taken for the coffee to cool from 80°C to 60°C at the same room temperature is \(\frac{13}{5}t\).