Answer
$9.15°C$
Work Step by Step
To calculate the temperature increase if the heat were not dissipated into the environment, we need to use the specific heat capacity of water.
Heat produced by the oxidation of 1 mole of glucose = 686 kcal
Specific heat capacity of water = 1 kcal/kg°C
Mass of water (body) = 75 kg
Temperature increase = Heat produced / (Specific heat capacity × Mass of water)
Temperature increase = 686 kcal / (1 kcal/kg°C × 75 kg) ≈ 9.15°C
