Final answer:
The man will metabolize 700 kJ of food energy to do 35.0 kJ of work at a 5.00% efficiency. An additional 665 kJ of heat will be transferred to the environment to maintain his body temperature.
Step-by-step explanation:
To calculate how much food energy a man will metabolize in the process of doing 35.0 kJ of work with an efficiency of 5.00%, we need to consider that efficiency (η) is the ratio of the work done (output) to the energy consumed (input). Given that η = (work done) / (energy consumed), we can rearrange this equation to solve for the energy consumed:
energy consumed = work done / efficiency
energy consumed = 35.0 kJ / 0.05 (5.00% as a decimal)
energy consumed = 700 kJ
Therefore, the man will need to metabolize 700 kJ of food energy. Regarding part (b) of the question, to determine the amount of heat transfer to the environment, we apply the first law of thermodynamics, which states that the change in internal energy is equal to the heat transfer minus the work done. If the man's temperature remains constant, then the change in internal energy is zero and all the consumed energy that is not converted into work is transferred as heat to the environment:
heat transfer = energy consumed - work done
heat transfer = 700 kJ - 35.0 kJ
heat transfer = 665 kJ
Thus, 665 kJ of heat would be transferred to the environment to keep his temperature constant.