Please assist with this Energy word problem

Question

Energy is conventionally measured in Calories as well as in Joules. One Calorie in nutrition is one kilocalories, defined as 1kcal=4186J. Metabolizing 1g of fat can release 9.00kcal. A student decided to try to lose weight by exercising. He plans to run up and down the stairs in a football stadium as facts as he can and as many times as necessary. To evaluate the program, suppose he runs up flight of 100 steps, each 0.30m high, in 105s For simplicity, ignore the energy he uses in coming down (which is small). Assume a typical efficiency for human muscles is 20%. This statement means that when your body converts 100J from metabolizing fat, 20J goes into doing mechanical work (here, climbing stairs). The remainder goes into extra internal energy. Assuming the students mass is 70kg:

a) How many times must he run the flight of stairs to lose two pounds of fat?

b) What is this average power output, in watts and in horsepower, as he is running up the stairs?

Answer 1

There are 453.6 grams in 1 pound (on Earth) so the student is planning on losing 907.2 grams of fat and since 1 gram of fat burns 9,00 kcal, the student must expend 8164.8 kcal or 34177853 joules. But that is at 100% efficiency. So at 20% efficiency, the work required would be 170889264 joules.

Work = F•d and F = weight = 70(9.8) = 686 N and the distance (up) is 30 m therefore the work to climb up once is 686•30 or 20580 joules.

The number of times required is (170889264 joules)/(20580 joules) or approx 8300 times

The power expended is energy/time = (20580 joules)/(105 s) = 196 watts. There are 745.7 wats per 1 hp so 196 watts is 0.263 hp