Question 1085510
{{{dT/dt=-k(T-A)}}}
Let's do a substitution,
{{{U=T-A}}}
{{{dU=dT}}}
So then,
{{{dU/U=-kdt}}}
{{{ln(U)=-kt+C}}}
{{{U=Ce^(-kt)}}}
{{{T-A=Ce^(-kt)}}}
and
{{{T(0)=68}}}
{{{T(5)=25}}}
So,
{{{68-20=Ce^(-k(0))}}}
{{{C=48}}}
and
{{{25-20=48e^(-5k)}}}
{{{5=48e^(-5k)}}}
{{{e^(-5k)=5/48}}}
{{{-5k=ln(5/48)}}}
{{{k=-ln(5/48)/5}}}
or approximately,
{{{k=-0.4524}}}
So,
{{{T-20=48e^(-0.4524t)}}}
Find t when T=21.
{{{21-20=48e^(-0.4524t)}}}
{{{e^(-0.4524t)=1/48}}}
{{{-0.4524t=ln(1/48){{{
{{{t=8.557}}}
or
{{{t=9}}}{{{hrs}}} to the nearest hour.