A liquid cools from70^(@)C to60^(@)C in 5 minutes. If the temperautre of the surrounding is 30^(@)C, what is the time taken by the liquid to cool from 50^(@)C to 40^(@)C ?

A liquid cools from70^(@)C to60^(@)C in 5 minutes. If the temperautre

1.	A liquid cools from70^(@)C to60^(@)C in 5 minutes. If the temperautre of the surrounding is 30^(@)C, what is the time taken by the liquid to cool from 50^(@)C to 40^(@)C ?
Answer» Solution :According to Newton.s law of cooling, Rate of COOLONG`prop` = {MEAN difference of twemperature between the body and the surroundings `(theta_(2)-theta_(1))/t prop((theta_(1)+theta_(2))/2)-theta_(0)` where t is the time TAKEN by the liquid to cool from `theta_(2)^(@)`C to `theta_(1)^(@)`C and `theta_(0)` is the TEMPERATURE of the surrounding `(60-50)/5 prop((60+50)/2)-30` `2 prop 55-30` `2 prop 25` ...(i) Let the time taken to cool from `50^(@)`C to `40^(@)`C be t, `(50-40)/t prop ((50+40)/2)-30` `10/(t) prop15` ...(ii) Dividing EQUATION (i) by (ii) `(2xxt)/10 = 25/(15) , t = (25xx10)/(2xx15)` = 8.33 s