A body cools from 80°C to 60° in 5 minutes. The temperature for the surrounding is 20°C. The time it takes to cool from 60°C to 40°C is

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Correct answer: C. (c) The formula to calculate the rate of cooling of the object is given by dT dt=K( T_f+T_i2-T_0) ...(a) For the first case, it can be written, using equation (a) that 80-605=K( 80+602-20) 4=50K...(b) If t is the required time for the second case, from equation (a), it can be written that 60-40t=K( 60+402-20) 20t=30K...(c) Divide equation (b) by equation (c) and solve to calculate the required time. 4 20t= 50K30K ⇒ t5= 53 ⇒t= 253 min=500s

$A body cools from 80°C to 60° in 5 minutes. The temperature for the surrounding is 20°C. The time it takes to cool from 60°C to 40°C is$

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