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The initial temperature of the coffee T0=100∘C.
The temperature of the room 15∘C.
Let T be the temperature at time t in minutes.
By Newton's law of cooling,
dtdTα(T−15)
∴dtdT=k(T−15)
⇒T−15dT=kdt
⇒∫T−15dT=∫kdt
⇒log(T−15)=kt+logc
⇒log(cT−15)=kt
⇒T−15=cekt
At t=0,T=100
We have 100−15=ce0⇒85=c
Thus we get T−15=85ekt.
When t=5minutes,T=60
⇒60−15=85e5k.
⇒e