an object kept in a large room having air temperature of 25°c takes 12min to cool from 80℃-70℃ the time taken to cool for the same object from70℃to 60℃would be nearly
Answers
the time taken to cool for the same object from 70° C to 60°C would be 7.5 min
explanation : Here, Initial temperature ( Ti) = 80°C
Final temperature ( Tf) = 70°C
Temperature of the surrounding ( To) = 25°C
time taken to cool down from 80° to 70°C , t = 12 min
A/C to Newton's law of cooling
Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To]
( Tf - Ti)/t = K[ ( 80 + 70)/2 - 25]
( 80-70)/12 = K[ 75 - 25]
5/3 = K × 50
K = 1/30 .......(1)
in second condition,
initial temperature ( Ti) = 70°C
Final temperature ( Tf) = 60°C
Time taken for cooling is t
A/C Newton's law of cooling
( 70 - 60)/t = K[ (70+60)/2 -25]
10/t = 1/30 × [65 - 25 ]
10/t = 40/30
t = 30/4 = 7.5 min
Answer:
15 minutes
Explanation:
first calculation A/C to Newton's law
dT/t =k {Ti + Tf /2} -To
or, 80-70/12 =k {80+70/2} - 25
or, k= 1/60
second calculation A/C to Newton's law
70-60/t = 1/60 {70+60/2} -25 ........(k=1/60)
or, t = 600/40 = 15 minutes