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Chapter 7: Problem 65

\(\bullet\) \(\bullet\) At 7.35 cents per kilowatt-hour, (a) what does it cost to operate a 10.0 hp motor for 8.00 hr? (b) What does it cost to leave a 75 W light burning 24 hours a day?

Short Answer

Expert verified
(a) $4.39 for the motor, (b) $0.13 for the light per day.

Step by step solution

01

Understanding the Units

First, we need to convert horsepower (hp) to watts (W) because the electricity cost is calculated in kilowatt-hours. One horsepower is equivalent to 746 watts. Thus, a 10 hp motor uses:\[10 \, \text{hp} \times 746 \, \text{W/hp} = 7460 \, \text{W}\] or 7.46 kilowatts (kW).
02

Calculate Energy Consumption for the Motor

Now, we find the total energy consumed by the 10 hp motor over 8 hours by multiplying its power in kW by the time in hours:\[7.46 \, \text{kW} \times 8 \, \text{hr} = 59.68 \, \text{kWh}\].
03

Calculate Cost for Operating the Motor

The cost of electricity is 7.35 cents per kilowatt-hour. To find the total cost, multiply the total energy consumption by the cost per kilowatt-hour:\[59.68 \, \text{kWh} \times 7.35 \, \text{cents/kWh} = 438.708 \, \text{cents}\]. Convert cents to dollars by dividing by 100:\[438.708 \, \text{cents} = 4.38708 \, \text{dollars}\].
04

Calculate Energy Consumption for the Light

A 75 W light bulb uses 0.075 kW. If it burns 24 hours a day, the energy consumption is:\[0.075 \, \text{kW} \times 24 \, \text{hr} = 1.8 \, \text{kWh/day}\].
05

Calculate Cost for Leaving the Light On

The cost to keep the 75 W light on for 24 hours is calculated by multiplying the energy consumption by the cost per kilowatt-hour:\[1.8 \, \text{kWh/day} \times 7.35 \, \text{cents/kWh} = 13.23 \, \text{cents/day}\]. Convert cents to dollars by dividing by 100:\[13.23 \, \text{cents/day} = 0.1323 \, \text{dollars/day}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horsepower to Watts Conversion
When dealing with motors or machines, power is often measured in horsepower (hp). However, when calculating energy costs, it's crucial to convert this unit to watts (W), as energy usage is typically measured in kilowatt-hours (kWh).
  • 1 horsepower is equivalent to 746 watts. This conversion factor is essential for converting the power of a 10 hp motor into watts.

For example, a 10 hp motor can be converted as follows: \[10 \times 746 = 7460 \text{ watts} = 7.46 \text{ kW}\] This conversion makes it easier to calculate energy consumption and the subsequent costs.
Energy Consumption
Energy consumption refers to the total amount of energy used by a device over a period. It plays a central role in determining the operating costs of electrical devices.
  • Energy consumption is often expressed in kilowatt-hours (kWh).
  • To find out how much energy an appliance consumes, multiply the power (in kW) by the time (in hours) the appliance is used.

For a 10 hp motor running for 8 hours:\[7.46 \text{ kW} \times 8 \text{ hours} = 59.68 \text{ kWh}\] This illustrates how much electricity the motor utilizes during this period, which is a critical step in calculating its operational cost.
Kilowatt-hour
A kilowatt-hour (kWh) is a measure of energy. It's one of the most common units used by electric companies to calculate electricity usage. Understanding this term is vital when it comes to energy billing.
  • One kilowatt-hour is the amount of electricity consumed by a device that uses 1,000 watts for one hour.
  • Appliances with higher energy consumption in watts will use more kWh over the same period.

For instance, if a 75 W light bulb is used continuously over 24 hours, it consumes:\[0.075 \text{ kW} \times 24 \text{ hours} = 1.8 \text{ kWh}\] This number helps us understand the electricity demand of a particular device per day.
Cost Per Kilowatt-hour
Electricity costs are typically expressed in terms of cost per kilowatt-hour (kWh). Knowing how to calculate this helps you predict and manage your energy costs effectively.
  • The energy provider sets the price per kWh, reflecting the cost to generate and supply electricity to your home or business.
  • To find the cost of electricity used, multiply the total kWh by the rate per kWh.

For example, if electricity costs 7.35 cents per kWh, running a 10 hp motor that consumes 59.68 kWh would cost:\[59.68 \text{ kWh} \times 7.35 \text{ cents/kWh} = 438.708 \text{ cents}\] Converting to dollars gives you an operation cost of \(4.38708 \text{ dollars}\).
Similarly, for a 75 W light burning all day:\[1.8 \text{ kWh/day} \times 7.35 \text{ cents/kWh} = 13.23 \text{ cents/day} = 0.1323 \text{ dollars/day}\]

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) A 250 g object on a frictionless, horizontal lab table is pushed against a spring of force constant 35 \(\mathrm{N} / \mathrm{cm}\) and then released. Just before the object is released, the spring is com- pressed 12.0 \(\mathrm{cm} .\) How fast is the object moving when it has gained half of the spring's original stored energy?

\(\bullet\) The engine of a motorboat delivers 30.0 \(\mathrm{kW}\) to the propeller while the boat is moving at 15.0 \(\mathrm{m} / \mathrm{s} .\) What would be the ten- sion in the towline if the boat were being towed at the same speed?

\(\bullet\) \(\bullet\) \(\bullet\) Automotive power. A truck engine transmits 28.0 \(\mathrm{kW}\) \((37.5 \mathrm{hp})\) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 \(\mathrm{km} / \mathrm{h}(37.7 \mathrm{mi} / \mathrm{h})\) on a level road. (a) What is the resisting force acting on the on a level road. (a) What is the resisting force acting on the truck? (b) Assume that 65\(\%\) of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 \(\mathrm{km} / \mathrm{h} ?\) At 120.0 \(\mathrm{km} / \mathrm{h} ?\) Give your answers in kilowatts and in horse- power.

\(\cdot\) \(\cdot\) Two tugboats pull a disabled supertanker. Each tug exerts a constant force of \(1.80 \times 10^{6} \mathrm{N}\) , one \(14^{\circ}\) west of north and the other \(14^{\circ}\) east of north, as they pull the tanker 0.75 \(\mathrm{km}\) toward the north. What is the total work they do on the supertanker?

\(\cdot\) It takes 4.186 \(\mathrm{J}\) of energy to raise the temperature of 1.0 \(\mathrm{g}\) of water by \(1.0^{\circ} \mathrm{C}\) (a) How fast would a 2.0 \(\mathrm{g}\) cricket have to jump to have that much kinetic energy? (b) How fast would a 4.0 \(\mathrm{g}\) cricket have to jump to have the same amount of kinetic energy?
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