Fill in the blanks:
$(a)$ $1 \ \text{Joule} = \dots \ \text{erg}$
$(b)$ $1 \ \text{eV} = \dots \ \text{Joule}$
$(c)$ $1 \ \text{kWh} = \dots \ \text{Joule}$

(N/A) Since $1 \ \text{J} = 1 \ \text{kg} \cdot \text{m}^2/\text{s}^2$ and $1 \ \text{erg} = 1 \ \text{g} \cdot \text{cm}^2/\text{s}^2$,we have $1 \ \text{J} = (10^3 \ \text{g}) \times (10^2 \ \text{cm})^2 / \text{s}^2 = 10^7 \ \text{erg}$.
$(b)$ Since $1 \ \text{eV}$ is the energy gained by an electron accelerated through a potential difference of $1 \ \text{V}$,$1 \ \text{eV} = (1.602 \times 10^{-19} \ \text{C}) \times (1 \ \text{V}) = 1.602 \times 10^{-19} \ \text{J}$.
$(c)$ Since $1 \ \text{kWh} = (10^3 \ \text{W}) \times (3600 \ \text{s}) = 1000 \ \text{J/s} \times 3600 \ \text{s} = 3.6 \times 10^6 \ \text{J}$.