Question:
Hot water cools from \(60^\circ\mathrm{C}\) to \(50^\circ\mathrm{C}\) in the first 10 minutes and to \(42^\circ\mathrm{C}\) in the next 10 minutes. The temperature of the surroundings is:
Hot water cools from \(60^\circ\mathrm{C}\) to \(50^\circ\mathrm{C}\) in the first 10 minutes and to \(42^\circ\mathrm{C}\) in the next 10 minutes. The temperature of the surroundings is:
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Apply Newton's law of cooling with the mean temperature in each interval and divide the two equations to eliminate the constant.
Updated On: Nov 11, 2026
- \(25^\circ\mathrm{C}\)
- \(10^\circ\mathrm{C}\)
- \(15^\circ\mathrm{C}\)
- \(20^\circ\mathrm{C}\)
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The Correct Option is B
Solution and Explanation
Newton's law (average form): \(\dfrac{\Delta T}{t} = k(T_{avg} - T_0)\).
Interval 1: \(\dfrac{60-50}{10} = k\left(\dfrac{60+50}{2} - T_0\right) \Rightarrow 1 = k(55 - T_0)\).
Interval 2: \(\dfrac{50-42}{10} = k\left(\dfrac{50+42}{2} - T_0\right) \Rightarrow 0.8 = k(46 - T_0)\).
Divide: \(\dfrac{1}{0.8} = \dfrac{55 - T_0}{46 - T_0}\) so \(1.25(46 - T_0) = 55 - T_0\).
\(57.5 - 1.25T_0 = 55 - T_0 \Rightarrow 2.5 = 0.25T_0 \Rightarrow T_0 = 10^\circ\mathrm{C}\).
Answer: (2) \(10^\circ\mathrm{C}\)
Interval 1: \(\dfrac{60-50}{10} = k\left(\dfrac{60+50}{2} - T_0\right) \Rightarrow 1 = k(55 - T_0)\).
Interval 2: \(\dfrac{50-42}{10} = k\left(\dfrac{50+42}{2} - T_0\right) \Rightarrow 0.8 = k(46 - T_0)\).
Divide: \(\dfrac{1}{0.8} = \dfrac{55 - T_0}{46 - T_0}\) so \(1.25(46 - T_0) = 55 - T_0\).
\(57.5 - 1.25T_0 = 55 - T_0 \Rightarrow 2.5 = 0.25T_0 \Rightarrow T_0 = 10^\circ\mathrm{C}\).
Answer: (2) \(10^\circ\mathrm{C}\)
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