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

Question

Vikas TU · Answer

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 = 2

in 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

Khimraj · Answer

According to Rate of cooling dT/dt = K((T_initial+T_final)/2 - T_surr) ( 80-60)/1 = K(70-60) K = 2 Time taken for cooling ( 60 - 50)/t = 2 ((60+50)/2 -60) t = 60 sec

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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

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

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