Foundations - Topic B - Energy - Exercises - Page 24: B.10(a)

$$128 \space J$$

Molar mass of Lead $(Pb)$ = $207.2 \space g/mol$ $\Delta T = 10.0 \space ^oC = 10.0 \space K$* * Only for temperature differences $(\Delta T)$ - Rearranging to do a dimensional analysis: $$100 \space g \times \frac{1 \space mol}{207.2 \space g} \times 10.0 \space K \times \frac{26.44 \space J}{\space K \space mol} = 128 \space J$$