Answer
(a) $-5.03\times10^{5}J$
(b) $120.16\space Calories$
Work Step by Step
(a) Here we use equations 15.1 $Q=\Delta U+W$ and 12.5 $Q=mL_{v}$ to find the change in internal energy of the weight lifter.
$Q=\Delta U+W=>\Delta U=Q-W=-mL_{v}-W$
Let's plug known values into this equation.
$\Delta U=-(150\space kg)(2.42\times10^{6}J/kg)-(1.4\times10^{5}J)$
$\Delta U=-5.03\times10^{5}J$ (change in internal energy of the weight lifter)
(b)
The number of nutritional = $(5.03\times10^{5}J)(\frac{1\space Calorie}{4186\space J})$
calories
The number of nutritional$=120.16\space Calories$
calories