# 5.25 calories are absorbed by an iron nail. How many joules is this?

By definition, $\text{1 cal}$ is the approximate amount of energy needed to raise the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$ at $\text{1 atm}$.
In thermochemistry, this is known to be equal to $\text{4.184 J}$ at ${25}^{\circ} \text{C}$. Therefore...
$5.25 \cancel{\text{cal" xx "4.184 J"/cancel"cal" = "21.97 J}}$
Or to three sig figs, $\approx$ $\underline{\textcolor{b l u e}{\text{22.0 J}}}$.