Question #515bb

1 Answer
yumean1119
Dec 8, 2017

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

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