Question

How long will the energy in a 1470-kJ (350-kcal) cup of yogurt last in a woman doing work at the rate of 150 W with an efficiency of 20.0% (such as in leisurely climbing stairs)? Does this time imply that it is easy to consume more food energy than you can reasonably expect to work off with exercise?

Answer 1

Answer:

The 1470 kJ of energy supplied by the yogurt would be consumed by 150 W of work (efficiency of 20%) in about 32.7 h. Then if people is not careful, easily would have more energy from the food that the amount that can consume exercising.

Explanation:

Step 1.

• Energy supplied: 1470 kJ = 1470000 J.

• Rate of energy used for the : 150 W = 150 J/s. This means that each second of this kind of work requieres 150 J of energy,

Step 2

For this case, efficiency is defined as the relationship between the actual rate of energy used, to the rate of energy consumed to obtain it. So:

$Eficiency=\frac{E_{obtain} }{E_{consumed} }$

Then the rate of energy required for this work, is calculated solving for $E_{Consumed}$:

$E_{consumed} =\frac{E_{obtain}}{Efficiency} =\frac{150\frac{J}{s} }{0.20} =750\frac{J}{s}$

This means that for a work that as a rate of 150 J/s , It requieres to consume energy at a rate of 750 J/s.

Step 3

Now, it is possible to calculate how much time would be needed to consume the 1470000 J of energy of the yogurt. It done using this relationship:

$Time=\frac{Energy}{rate}$

Where Energy stands for the total amount of energy that would be consumed, and rate stans for the rate of energy consumption, Then the time would be:

$Time=\frac{1470000J}{750\frac{J}{s} }=1960s=32.7 h$

Then if for the consumption of 1470 kJ of energy is required almost 33 hours, Then is easy to see that the amount of enegy that can be obtain of food would much more than the amount that can be depleted by exercise