Answer
The yearly rate of change of $T$ is $\dfrac{dT}{dt}$.
$\dfrac{dT}{dt}=0.0972674763$ K/year
Work Step by Step
The yearly rate of change of $T$ is $\dfrac{dT}{dt}$.
To find $\dfrac{dT}{dt}$.
Firstly calculate derivate of $R = σT^{4}$ with respect to time $t$ using the chain rule.
We get, $\dfrac{dR}{dt}=\dfrac{dR}{dT}\times \dfrac{dT}{dt}$
Now derivate $R = σT^{4}$ with respect to temperature $T$ using power rule.
We get, $\dfrac{dR}{dT}=4\sigma T^{3}$
Substitute $\dfrac{dR}{dT}=4\sigma T^{3}$ in $\dfrac{dR}{dt}=\dfrac{dR}{dT}\times \dfrac{dT}{dt}$.
We get $\dfrac{dR}{dt}=4\sigma T^{3}\times \dfrac{dT}{dt}$
Now solve for $\dfrac{dT}{dt}$.
We get, $\dfrac{dT}{dt}=\dfrac{1}{4\sigma T^{3}}\times\dfrac{dR}{dt}$
Substitute $\dfrac{dR}{dt}=0.5 Js^{−1}m^{−2} per\hspace{4px} year.$, $T=283 K$ and $σ = 5.67 ×10^{−8} Js^{−1}m^{−2}K^{−4}$.
$\dfrac{dT}{dt}=\dfrac{1}{4\times5.67\times10^{-8}\times 283^{3}}\times0.5=0.0972674763$ K/year
