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A cup of tea cool from 80 degree Celsius to 60 degree celsius in one minute the abandoned temperature is 30 degree Celsius in cooling from 60 degree Celsius to 50 degree Celsius it will take
Dear student Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To] ( Tf - Ti)/t = K[ ( 80 + 60)/2 - 60]( 80-60)/1 = K[ 70 - 60]20 = K× 10 K = 20/10 = 2in second condition, initial temperature ( Ti) = 60°C Final temperature ( Tf) = 50°C Time taken for cooling is t A/C Newton's law of cooling ( 60 - 50)/t = 2 [ (60+50)/2 -60]10/t = 2* 5 10/t = 10 t = 1 min
According to Rate of coolingdT/dt = K((T_initial+T_final)/2 - T_surr) ( 80-60)/1 = K(70-60)K = 2Time taken for cooling(60 - 50)/t = 2 ((60+50)/2 -60)t = 60 sec
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