JEE Mains · Physics · STD 11 - 10.2 transmission of heat
A body cools in \(7\) minutes from \(60^{\circ}\,C\) to \(40^{\circ}\,C\). The temperature of the surrounding is \(10^{\circ}\,C\). The temperature of the body after the next \(7\) minutes will be
- A \(32^{\circ}\,C\)
- B \(30^{\circ}\,C\)
- C \(28^{\circ}\,C\)
- D \(34^{\circ}\,C\)
Answer & Solution
Correct Answer
(C) \(28^{\circ}\,C\)
Step-by-step Solution
Detailed explanation
using average rate of Newton's law of cooling \(\frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_s\right)\) Given \(\frac{60-40}{7}=K(50-10) \ldots\) (i) \(\frac{40-T}{7}=K\left(\frac{40+T}{2}-10\right) \ldots\) (ii) From \((i)\) and \((ii)\) \(T =28^{\circ}\,C\)
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