law of cooling

JEE Main 2015 Discussion Forum › Thermal physics questions › law of cooling

Temperature of a body comes down to 40Â°C from 50Â°C in 5 min...in how much time temperature will come down to 30Â°C...temp of surroundings 20Â°C..

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#Physics #Thermal physics

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7 Answers

11

virang1 Jhaveri ·2009-04-16 01:18:24

10/5 = k(30)
k= 1/15
10/t=k*(20)
1/2t =1/15
t = 7.5mins

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-04-16 01:21:07

ans. 8.5 mins

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1

Aditya ·2009-04-16 12:34:25

What Virang has done should be correct!

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1

Â°áƒ¦â€¢à¹“ÑÏ…Î ·2009-04-16 12:45:00

rate of coolin =k(t-ts)
here t=(avg temp over d intrval)

so
10/5 =k(45-20)

10/t=k(35-20)

u get ur answer :)

cheero~

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1

skygirl ·2009-04-16 20:34:26

no.. watever mrunal has done is correct ..

take the average tem and find its diff from the room temp surrounding temp. then apply newton's law [that cololing wala formula]

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-04-17 09:16:22

should not we use that log ..blah..blah...wala formula???[7]

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3

iitimcomin ·2009-04-22 10:50:26

abhirup this is a safe approximation used in such situations .............. :)

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