As per Newton's Law of Cooling, the rate of the cooling is always proportional to the temperature difference between the object and its surroundings.
dtdT=−k(T0−Ts)
Tt=Ts+(T0−Ts)e−kt
T0=20°C And Ts=5°C
At t=60, T60=12°C
12=5+(20−5)e−60k
12−5=15e−60k
157=e−60k
k=−60ln(157)
k=0.0127023342
a)
After another one minute, ∴t=120s
Tt=Ts+(T0−Ts)e−kt
T120=5+(15)e−0.0127023342∗120
=5+0.01611786424
=5.01611786424°C
b)
When temperature will read 6°C
Tt=Ts+(T0−Ts)e−kt
6=5+(15)e−0.0127023342t
1=(15)e−0.0127023342t
e−0.0127023342t=151=0.06666666667
t=−0.0127023342ln0.06666666667
t=213.19s
=3.55minutes
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