Two resistors of 6 and 8 ohms each are linked in series to a 12 V battery in an electrical circuit. How much heat will the 8 Ω resistor dissipate in 5 seconds?

Resistance $\left(R_1\right)=6\Omega$
Resistance $\left(R_2\right)=8\Omega$
Potential difference $\left(V\right)=12 \mathrm{V}$
Time $\left(t\right)=5 \mathrm{s}$
Resistors are connected in series and therefore, the total resistance is:
 ${\mathrm{R}}_{\mathrm{eq}}=6+8=14\mathrm{\Omega }$
The total current is calculated using Ohm's law as follows,
$\mathrm{I}=\left(\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{eq}}}\right)$
$\Rightarrow \mathrm{I}=\left(\frac{12}{14}\right)=0.85 \mathrm{A}$
Hence, the heat dissipated by $8\Omega$ in 5 seconds will be
$\mathrm{H}={\mathrm{I}}^2{\mathrm{R}}_2\mathrm{t}=0.{85}^2\times 8\times 5=28.9 \mathrm{J}$

Therefore, 8 Ω resistor will dissipate 28.9 J of heat in 5 seconds.