# You decide to use your body as a Carnot heat engine. The operating gas is in a tube with one end in your mouth (where the temperature is 37.0 C) and the other end at the surface of your skin, at 30.0 C.

What is the maximum efficiency of such a heat engine?
e=2.26%

Suppose you want to use this human engine to lift a 2.05 kg box from the floor to a tabletop 1.20 m above the floor. How much must you increase the gravitational potential energy?
W=24.1 J

Suppose you want to use this human engine to lift a 2.05 kg box from the floor to a tabletop 1.20 m above the floor. How much heat input is needed to accomplish this?
Q=1070 J

How many 350 calorie (those are food calories, remember) candy bars must you eat to lift the box in this way? Recall that 80.0 % of the food energy goes into heat.
??? candy bars

*I know that it will be a very small number, but I'm not quite sure how to calculate it!

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1. Carnot efficiency = 1 - 303/310 = 2.26% OK so far

W = 2.05 kg*9.8 m/s^2*1.2 m = 24.1 J Right again.

0.0226Q = 24.1 J
Q = 1066 J close enough

1066 J is 255 calories but that is 0.255 Kcal, or Food Calories. Each candy bar is 350,000 (gram)-calories. Eating one bar produces 0.8*350,000 cal of heat or 280,000 cal.

You only need to consume 255/280,000 = 0.0009 of the candy bar.

This is a self-contradicting problem because in the beginning it says to treat the body as a 2.26% efficient engine but in the end it says to assume 20% efficiency (80% going to heat) when calculating the candy bar requirement. The body is able to extract more than Carnot efficiency from food because it is NOT a heat engine. There are other less entropy-producing ways to convert chemical energy to work - electrochemical processes. That is how fuel cells work

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