A body cools in 7 min from $60^{\circ}C$ to $40^{\circ}C$. What times (in min) does it take to cool from $40^{\circ}C$ to $28^{\circ}C$, if surrounding temperature is $10^{\circ}C$? (Assume Newton's law of cooling)}
Show Hint
- 3.5
- 14
- 7
- 10
The Correct Option is C
Solution and Explanation
Newton's law of cooling: $\frac{\theta_1 - \theta_2}{t} = K\left(\frac{\theta_1 + \theta_2}{2} - \theta_0\right)$.
Step 2: Detailed Explanation:
For first case: $\frac{20}{7} = K(50 - 10) = 40K$. For second case: $\frac{12}{t} = K(34 - 10) = 24K$. Dividing, $\frac{20/7}{12/t} = \frac{40K}{24K} \Rightarrow \frac{20t}{84} = \frac{5}{3} \Rightarrow t = 7$ min.
Step 3: Final Answer:
The time taken is 7 minutes.
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