A liquid contained in a calorimeter cools from 80°C to 60°C in 10s.

1.	A liquid contained in a calorimeter cools from 80°C to 60°C in 10s. If the temperature of thesurrounding is 30°C, calculate the time it will take to cool further to 45'C
Answer» According to Newton's law of cooling, we havemc (T1- T2)/t = k [(T1+ T2)/2 - T0)] ---------------(1)mc/k = [(T1+ T2)/2 - T0)]/[(T1- T2)/t]mc/k =[(80+50)/2 - 20)]/[(80- 50)/5] = 45/6Substituting T1 = 60 and T2 = 30 and T0 = 20 (all in°C) in Eq. (1), we getmc (60- 30)/t = k [(60+ 30)/2 - 20)]30 mc/t = 25 kt = 30 mc/(25k) = (30/25) x (45/6) = 9 minThe body takes 9 min to cool from60°Cto30°C. but the answer is 13.5 seconds

A liquid contained in a calorimeter cools from 80°C to 60°C in 10s. If the temperature of thesurrounding is 30°C, calculate the time it will take to cool further to 45'C

Answer»

According to Newton's law of cooling, we havemc (T1- T2)/t = k [(T1+ T2)/2 - T0)] ---------------(1)mc/k = [(T1+ T2)/2 - T0)]/[(T1- T2)/t]mc/k =[(80+50)/2 - 20)]/[(80- 50)/5] = 45/6Substituting T1 = 60 and T2 = 30 and T0 = 20 (all in°C) in Eq. (1), we getmc (60- 30)/t = k [(60+ 30)/2 - 20)]30 mc/t = 25 kt = 30 mc/(25k) = (30/25) x (45/6) = 9 minThe body takes 9 min to cool from60°Cto30°C.

but the answer is 13.5 seconds

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A liquid contained in a calorimeter cools from 80°C to 60°C in 10s. If the temperature of thesurrounding is 30°C, calculate the time it will take to cool further to 45'C

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