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Grade: 12

                        

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

one year ago

Answers : (2)

Vikas TU
13785 Points
							
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
one year ago
Khimraj
3007 Points
							
 
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
one year ago
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