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JEE Mains · Physics · STD 11 - 10.2 transmission of heat

એક પદાર્થ \(5\) મિનિટમાં \(80^{\circ}\,C\) માથી \(60^{\circ}\,C\) સુધી ઠંડો પડે છે.પરિસરનું તાપમાન \(20^{\circ}\,C\) છે.તો તેને \(60^{\circ}\,C\) થી \(40^{\circ}\,C\) સુધી ઠંડો પડવા માટેનો સમય .......... \(s\) થશે.

  1. A \(500\)
  2. B \(\frac{25}{3}\)
  3. C \(450\)
  4. D \(420\)
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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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