Question:
Sitha prepared tea for her brother. After preparing tea the temparature of the tea was 80 degree celsius. She kept it for 5 minutes and gave to her brother when the temparature reached 50 degree celcius . If the surrounding temparature is 20 degree celcius.Calculate the time it takes to cool from 60 degree celcius to 30 degree celcius.
Answer:
Let us consider two cases here. Case 1 – Temperature changing from 800C to 500C ; Case 2 – Temperature changing from 600C to 300C
According to Newton’s law of cooling, dT/dt = -K(T – T0) where dT/dt is the change in temperature with time dt; T is the average temperature of the hot body when it falls from temperature T1 to T2
Case 1
Change in temperature dT =800C – 500C = 300C
Average temperature T = (80 + 50) /2 = 650C
Time of fall = 5 min
Temperature of the surroundings T0 = 200C
In the above case, Case 1, dT/dt = -K(T – T0)
30/5 = -K(65 – 20)
-K = 6/45 = 2/15
Case 2
Change in temperature dT =600C – 300C = 300C
Average temperature T = (60 + 30) /2 = 450C
Time of fall = t min
Temperature of the surroundings T0 = 200C
In the above case, Case 1, dT/dt = -K(T – T0)
30/t = -K(45 – 20)
We know K = -2/15 from case 1
Hence, 30/t = (2/ 15)* (25)
T = 9 min
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