A container of hot water at 80°C cools to 79°C in 15 seconds when it is placed in a room that is at 20°C. Use

Question

A container of hot water at 80°C cools to 79°C in 15 seconds when it is placed in a room that is at 20°C. Use Newton's law of cooling to estimate the time it will take for the container to cool from 60°C to 59°C.

Answer 1

dT/dt = - k (T-To)
dT / (T-To) = - k dt
log (T-To) = - k t + c ……(1)
at two times there will 2 temps

log (T1-To) = - k t1 + c
log (T2-To) = - k t2 + c

log (T2-To) - log (T1-To)= - k (t2-t1)

k = - {1/(t2 - t1)} log [(T2-To) / (T1-To)] ----(2)

To = 20
t2-t1 =15 , T1=80
T2 =79
k = - (1/15) log [59/60]
k = (1/15) log [59/60] =0.00112
again
To = 20
t2-t1 =t , T1=60
T2 =59
k = 0.00112

(t2 - t1) = - {1/k} log [39 / 40]

(t2 - t1) = - {1/0.00112} { - 0.025317}

time taken = 22.60 secs