Question:
A body cools from \(100^{\circ}\text{C}\) to \(60^{\circ}\text{C}\) in 20 minutes, the temperature of the surroundings being \(20^{\circ}\text{C}\). The total time taken (in minutes) for the body to cool down to \(40^{\circ}\text{C}\) is ...
A body cools from \(100^{\circ}\text{C}\) to \(60^{\circ}\text{C}\) in 20 minutes, the temperature of the surroundings being \(20^{\circ}\text{C}\). The total time taken (in minutes) for the body to cool down to \(40^{\circ}\text{C}\) is ...
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Temperature difference from the surroundings decays exponentially.
Updated On: Oct 1, 2026
- \(60\)
- \(40\)
- \(50\)
- \(30\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
By Newton law of cooling, \(\frac{dT}{dt} = -k(T - T_s)\), so \(T - T_s = (T_0 - T_s)e^{-kt}\). Here \(T_s = 20\) C and \(T_0 = 100\) C.
Step 2: Key Formula or Approach:
Excess temperature: initially \(80\) C. After \(20\) min it is \(60 - 20 = 40\) C.
Step 3: Detailed Explanation:
\(40 = 80e^{-20k}\), so \(e^{-20k} = \frac12\) and \(k = \frac{\ln 2}{20}\).
To reach \(40\) C the excess is \(20\) C: \(20 = 80e^{-kt}\), so \(e^{-kt} = \frac14\), i.e. \(kt = \ln 4 = 2\ln 2\).
\[ t = \frac{2\ln 2}{\ln 2/20} = 40\ \text{min} \]
Final Answer:
The body takes \(40\) minutes in total, option (B). \[ \boxed{40} \]
By Newton law of cooling, \(\frac{dT}{dt} = -k(T - T_s)\), so \(T - T_s = (T_0 - T_s)e^{-kt}\). Here \(T_s = 20\) C and \(T_0 = 100\) C.
Step 2: Key Formula or Approach:
Excess temperature: initially \(80\) C. After \(20\) min it is \(60 - 20 = 40\) C.
Step 3: Detailed Explanation:
\(40 = 80e^{-20k}\), so \(e^{-20k} = \frac12\) and \(k = \frac{\ln 2}{20}\).
To reach \(40\) C the excess is \(20\) C: \(20 = 80e^{-kt}\), so \(e^{-kt} = \frac14\), i.e. \(kt = \ln 4 = 2\ln 2\).
\[ t = \frac{2\ln 2}{\ln 2/20} = 40\ \text{min} \]
Final Answer:
The body takes \(40\) minutes in total, option (B). \[ \boxed{40} \]
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