Question

what is the efficiency of an out-of-condition professor who does 1.90 ✕ 105 j of useful work while metabolizing 490 kcal of food energy?

Answer 1

The efficiency of the out-of-condition professor is approximately 0.09%.

Step-by-step explanation:

The efficiency of the out-of-condition professor can be calculated using the formula:

Efficiency = (Useful work / Energy input) x 100%

Given that the professor does 1.90 x 10^5 J of useful work and metabolizes 490 kcal of food energy, we need to convert the food energy to joules first. One kilocalorie (kcal) is equal to 4184 joules, so 490 kcal is equal to 490 x 4184 J.

Now we can calculate the efficiency:

Efficiency = (1.90 x 10^5 J / (490 x 4184 J)) x 100%

Efficiency ≈ 0.09%