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A body cools from 60^@C to 50^@C in 10 min of room If the room temperature is25^@ C and assuming Newton's cooling law holds good, the temperature of the body after 10 more minute. |
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Answer» Solution :ACCORDING to newtons law ofcooling ` ( theta_2 - theta_1)/(t) = k [ (theta_2 + theta_1)/(2) - theta_0 ]` ` therefore (60 - 50)/(10)= k [ (60 + 50)/(2) - 25]` ` 1 = 30 k " or" k = 1/30` ` (50 - theta)/(10) = k [ (50 +theta)/(2) - 25 ] = k theta / 2 = (theta)/(60)` or ` 70 theta =3000, theta = 300/7 = 42.8^@C` |
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