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 the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

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

Hint:

Use Newton's law of cooling to solve such problems. The temperature difference between the object and the environment is proportional to the rate of cooling.

Answer 1

The Correct Option is A

By using the average form of Newton's law of cooling: \[ \frac{90 - 80}{t} = k \left( \frac{90 + 80 - 20}{2} \right) \quad \cdots (i) \] \[ \frac{80 - 60}{t'} = k \left( \frac{80 + 60 - 20}{2} \right) \quad \cdots (ii) \] From equations (i) and (ii), we get: \[ 10 \times t' = 65 \times t \] \[ t' = \frac{65}{50} \times 2t = \frac{13}{5} t \] Thus, the time taken to cool from 80°C to 60°C is \( \frac{13}{5} t \), so the correct answer is option (1).

Answer 2

Step 1: Apply Newton’s Law of Cooling.
According to Newton’s law of cooling, the rate of cooling is proportional to the temperature difference between the object and the surroundings.
\[ \frac{dT}{dt} = -k (T - T_s) \] where:
\( T \) = temperature of the body,
\( T_s \) = surrounding temperature,
\( k \) = proportionality constant.

Step 2: Integrate the equation.
Integrating between initial temperature \( T_1 \) and final temperature \( T_2 \):
\[ \int_{T_1}^{T_2} \frac{dT}{T - T_s} = -k \int_0^t dt \] \[ \ln \frac{T_2 - T_s}{T_1 - T_s} = -k t \] \[ k = \frac{1}{t} \ln \frac{T_1 - T_s}{T_2 - T_s} \]

Step 3: Case 1 — Cooling from 90°C to 80°C.
For the first cooling period:
\[ T_s = 20^\circ C, \, T_1 = 90^\circ C, \, T_2 = 80^\circ C \] Hence,
\[ k = \frac{1}{t} \ln \frac{90 - 20}{80 - 20} = \frac{1}{t} \ln \frac{70}{60} = \frac{1}{t} \ln \frac{7}{6}. \]

Step 4: Case 2 — Cooling from 80°C to 60°C.
Let the time taken for this cooling be \( t_2 \).
Using the same formula:
\[ k = \frac{1}{t_2} \ln \frac{80 - 20}{60 - 20} = \frac{1}{t_2} \ln \frac{60}{40} = \frac{1}{t_2} \ln \frac{3}{2}. \] Since \( k \) is the same for both cases, we equate the two expressions for \( k \):
\[ \frac{1}{t} \ln \frac{7}{6} = \frac{1}{t_2} \ln \frac{3}{2}. \] \[ t_2 = t \frac{\ln(3/2)}{\ln(7/6)}. \]

Step 5: Simplify the ratio.
Using logarithmic approximation or calculator values:
\[ \ln(3/2) = 0.405, \quad \ln(7/6) = 0.154. \] \[ \frac{\ln(3/2)}{\ln(7/6)} = \frac{0.405}{0.154} \approx 2.63 \approx \frac{13}{5}. \] \[ t_2 = \frac{13}{5} t. \]

Final Answer:
\[ \boxed{t_2 = \frac{13}{5}t} \]