Answer
Energy is dissipated in $R_2$ at a rate of $~~0.050~W$
Work Step by Step
Let $i_1$ be the current through $R_1$ and let's assume the direction of the current is to the right.
Let $i_2$ be the current through $R_2$ and let's assume the direction of the current is to the left.
Let $i_3$ be the current through $R_3$ and let's assume the direction of the current is down.
Using the junction rule:
$i_3 = i_1+i_2$
Using the loop rule counterclockwise around the right loop:
$1.00-2.00~i_2-5.00~i_3 = 0$
$1.00-2.00~i_2-5.00~(i_1+i_2) = 0$
$1.00-7.00~i_2-5.00~i_1 = 0$
$-5.00+35.0~i_2+25.0~i_1 = 0$
Using the loop rule clockwise around the left loop:
$3.00-4.00~i_1-5.00~i_3 = 0$
$3.00-4.00~i_1-5.00~(i_1+i_2) = 0$
$3.00-9.00~i_1-5.00~i_2 = 0$
$21.0-63.0~i_1-35.0~i_2 = 0$
We can add these two equations to find $i_1$:
$16.0-38.0~i_1 = 0$
$i_1 = \frac{16.0}{38.0}$
$i_1 = 0.421~A$
The current through resistance 1 is $~~0.421~A$
We can find $i_2$:
$1.00-7.00~i_2-5.00~i_1 = 0$
$1.00-7.00~i_2-(5.00)~(0.421) = 0$
$7.00~i_2 = -1.105$
$i_2 = \frac{-1.105}{7.00}$
$i_2 = -0.158~A$
The negative sign indicates that the current flows through $R_2$ to the right.
We can find the rate at which energy is dissipated in $R_2$:
$P = i_2^2~R_2$
$P = (0.158~A)^2(2.00~\Omega)$
$P = 0.050~W$
Energy is dissipated in $R_2$ at a rate of $~~0.050~W$
