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

This question involves the concept of Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature. The formula used for Newton's Law of Cooling is:

\(\frac{dT}{dt} = -k(T - T_R)\)

where:

• \(T\) is the temperature of the object at time \(t\) 
• \(T_R\) is the ambient (room) temperature
• \(k\) is a constant

The solution proceeds in two stages:

1. First cooling from 90°C to 80°C.
2. Then cooling from 80°C to 60°C.

Step 1: Cooling from 90°C to 80°C

Given that the coffee initially cools from 90°C to 80°C in \(t\) minutes. Applying the formula for Newton's Law of Cooling, the temperature falls from \(T_1 = 90°C\) to \(T_2 = 80°C\).

The average temperature \(T_{\text{avg}}\) during this time is:\)

\(T_{\text{avg}} = \frac{90 + 80}{2} = 85°C\)

Hence, the differential cooling equation becomes:

\(\frac{80 - 90}{t} = -k(85 - 20)\)

Simplifying gives:

\(-10 = -65kt\)

\(k = \frac{1}{6.5t}\)

Step 2: Cooling from 80°C to 60°C

Given the same room temperature, we calculate similarly where the temperature falls from \(T_1' = 80°C\) to \(T_2' = 60°C\). The average temperature \(T_{\text{avg}}'\) is:\)

\(T_{\text{avg}}' = \frac{80 + 60}{2} = 70°C\)

Using the differential cooling equation:

\(\frac{60 - 80}{t'} = -k(70 - 20)\)

Solving gives:

\(-20 = -50kt'\)

k = \frac{1}{6.5t}, we find:

t' gives:

\(t' = \frac{13}{5}t\)

Thus, the time taken by a similar cup of coffee to cool from 80°C to 60°C, while at the room temperature of 20°C, is \(\frac{13}{5}\) tim\)

The correct option is: \(\frac{13}{5}t\).