Answer
(a) $-492KJ$
(b) $117Cal$
Work Step by Step
(a) We can find the required internal energy as follows:
$\Delta U=Q-W$
$\implies \Delta U=-mL-W$
We plug in the known values to obtain:
$\Delta U=-(0.110)(2.26\times 10^6)-2.43\times 10^5$
$\Delta U=-492KJ$
(b) We can find the number of nutritional calories as
$\Delta U=(492KJ)(0.239Kcal/KJ)=117Cal$
