Question:
A body cools according to Newton's law of cooling from \(100^{\circ}\)C to \(60^{\circ}\)C in \(20\) minutes. The temperature of the surroundings being \(20^{\circ}\)C, then the total time required for the body to cool down to \(30^{\circ}\)C is
A body cools according to Newton's law of cooling from \(100^{\circ}\)C to \(60^{\circ}\)C in \(20\) minutes. The temperature of the surroundings being \(20^{\circ}\)C, then the total time required for the body to cool down to \(30^{\circ}\)C is
Show Hint
Use \(\ln\frac{T-T_s}{T_0-T_s} = -kt\) with surroundings at 20 degrees C.
Updated On: Oct 1, 2026
- \(90\) minutes
- \(1\) hour and \(10\) minutes
- \(80\) minutes
- \(60\) minutes
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
Newton's law of cooling: \(\frac{dT}{dt} = -k(T - T_s)\), so \(T - T_s = (T_0 - T_s)e^{-kt}\). The excess temperature falls exponentially.
Step 2: Find k:
At \(t = 20\) min: \(60 - 20 = (100-20)e^{-20k}\), so \(40 = 80e^{-20k}\), giving \(e^{-20k} = \frac12\), and \(20k = \ln2\).
Step 3: Find the time for 30 degrees:
We need \(30 - 20 = 80e^{-kt}\), so \(e^{-kt} = \frac18\) and \(kt = \ln8 = 3\ln2\).
Since \(20k = \ln2\), we get \(t = 3\times20 = 60\) minutes.
Step 4: Check:
The excess temperature halves every \(20\) minutes: \(80\to40\to20\to10\). The final excess is \(30 - 20 = 10\), reached after three halvings, i.e. \(60\) minutes. Options of \(80\) and \(90\) minutes are too long, and \(70\) minutes does not match three halvings.
Final Answer:
The total time is \(60\) minutes, option (D). \[ \boxed{60\ \text{minutes}} \]
Newton's law of cooling: \(\frac{dT}{dt} = -k(T - T_s)\), so \(T - T_s = (T_0 - T_s)e^{-kt}\). The excess temperature falls exponentially.
Step 2: Find k:
At \(t = 20\) min: \(60 - 20 = (100-20)e^{-20k}\), so \(40 = 80e^{-20k}\), giving \(e^{-20k} = \frac12\), and \(20k = \ln2\).
Step 3: Find the time for 30 degrees:
We need \(30 - 20 = 80e^{-kt}\), so \(e^{-kt} = \frac18\) and \(kt = \ln8 = 3\ln2\).
Since \(20k = \ln2\), we get \(t = 3\times20 = 60\) minutes.
Step 4: Check:
The excess temperature halves every \(20\) minutes: \(80\to40\to20\to10\). The final excess is \(30 - 20 = 10\), reached after three halvings, i.e. \(60\) minutes. Options of \(80\) and \(90\) minutes are too long, and \(70\) minutes does not match three halvings.
Final Answer:
The total time is \(60\) minutes, option (D). \[ \boxed{60\ \text{minutes}} \]
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