Chapter 14, Problem 4P

An average active person consumes about 2500 Cal a day. (a) What is this in joules? (b) What is this in kilowatt-hours? (c) If your power company charges about 10 $ per kilowatt-hour, how much would your energy cost per day if you bought it from the power company? Could you feed yourself on this much money per day?

To determine

Part(a)To determine:

To convert energy in calories to joules.

Answer to Problem 4P

Solution:

2500 Calories is $\text{0}\text{.00290556}$ kWhr

Explanation of Solution

Formula Used:

1000 Calories is 4186 J

Calculation:

If 1000 Calories is 4186 J, then 1 Calories is $\frac{4186}{1000}=4.186$ J

Then 2500 Calories must be $2500\frac{4186}{1000}=10465$ J

To determine

Part(b)To determine:

To convert energy in calories to KiloWatt-Hours.

Answer to Problem 4P

Solution:

2500 Calories is 10465 KWhr

Explanation of Solution

Formula Used:

1000 Calories is 0.00116222KWhr

Calculation:

If 1000 Calories is 0.00116222 KWhr, then 1 Calories is $\frac{0.00116222}{1000}$ KWhr

Then 2500 Calories must be $2500\frac{0.00116222}{1000}=\text{0}\text{.00290556}$ KWhr

To determine

Part(c)To determine:

Cost for energy consumption.

Answer to Problem 4P

Solution:

Energy cost per day is $\text{0}\text{.0290556}$ cents

Explanation of Solution

Formula Used:

1000 Calories is 0.00116222 KWhr

Calculation:

If 1000 Calories is 0.00116222 KWhr, then 1 Calories is $\frac{0.00116222}{1000}$ KWhr

Then 2500 Calories must be $2500\frac{0.00116222}{1000}=\text{0}\text{.00290556}$ KWhr

If it costs 10 cents per KWhr, then it would cost $(10)(\text{0}\text{.00290556)=0}\text{.0290556}$ cents