Q$1$ b: An initial temperature of cylinder at $1200K$ is allowed to cool down in air at $300K.$ The differential equation for the temperature of the cylinder is given,

$\frac{d}{dt}=-2\mathrm{.}2067{10}^{-12}({}^4-81{10}^8)$

Where $$ is in $K$ and $t$ in seconds. Find the temperature at $360$ second. Assume a step size of $120$ second.

I want a paper solution, not typing on the keyboard."modeling"