Question

Question #515bb

Answer 1

About `147`F

Explanation:

The differential equation in this model is follows.
`(dT)/dt = -K(T-T_(env))`, where
`T`: temparature of the body
`T_(env)`: temparature of the environment
`t`: time
`K`: time constant

Now, let's solve this.

[1] Let `T'=T-T_(env)`
`(dT')/dt=-KT'`

[2] Separate variables:
`(dT')/(T')=-Kdt`

[3] Integrate both sides;
`ln T'=-Kt+C` (C:constant)
`T'=e^(-Kt+C)`

Therefore, the result is
`T=T_(env)+e^(-Kt+C)`

Plug in the given value.(`T_(env)=76, K=0.03`
When `t=0, T=172`:
`172=76+e^C`
`e^C=96`
`T=76+96e^-0.03t`

When `t=10`
`T=76+96*e^-(0.03*10)=147.1185…`

Here are some useful tips:
https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/exponential-models-diff-eq/v/newtons-law-of-cooling