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JEE Mains · Physics · STD 11 - 10.2 transmission of heat
A body cools from \(80^{\circ}\,C\) to \(60^{\circ}\,C\) in \(5\) minutes. The temperature of the surrounding is \(20^{\circ} C\). The time it takes to cool from \(60^{\circ}\,C\) to \(40^{\circ}\,C\) is........... \(s\)
- A \(500\)
- B \(\frac{25}{3}\)
- C \(450\)
- D \(420\)
Answer & Solution
Correct Answer
(A) \(500\)
Step-by-step Solution
Detailed explanation
Rate of cooling \(\alpha\) Temperature difference \(\frac{80-60}{5}=k\{70-20\}\) \(\frac{60-40}{t}=k[50-20]\) \(\frac{4 t}{20}=\frac{50}{30}\) \(t=\frac{25}{3} min =500\,sec\) \(\Rightarrow t=500 \text { seconds }\)
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