/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 If a kitchen-sink water tap leak... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If a kitchen-sink water tap leaks one drop per second, how many gallons of water are wasted annually? What is the mass of the wasted water, in lb? Assume that there are 46,000 drops per gallon and that the density of water is \(62.3 \mathrm{lb} / \mathrm{ft}^{3}\).

Short Answer

Expert verified
The water wasted annually is 685.57 gallons and the mass of the wasted water is 5,709.55 lb.

Step by step solution

01

- Calculate total drops in a year

First, find out how many seconds are in a year. There are 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, and 365 days in a year. Therefore, the number of seconds in a year is calculated as follows: \( 60 \times 60 \times 24 \times 365 = 31,536,000 \text{ seconds/year} \) So, the faucet leaks 31,536,000 drops in a year.
02

- Convert drops to gallons

Given that there are 46,000 drops per gallon, we can find the number of gallons wasted annually by dividing the total number of drops by the number of drops per gallon: \( \frac{31,536,000 \text{ drops/year}}{46,000 \text{ drops/gallon}} = 685.57 \text{ gallons/year} \)
03

- Convert gallons to cubic feet

To find the volume in cubic feet, use the conversion that 1 gallon is equivalent to 0.133681 cubic feet: \( 685.57 \text{ gallons/year} \times 0.133681 \text{ ft}^3/\text{gallon} = 91.69 \text{ ft}^3/\text{year} \)
04

- Calculate the mass of the wasted water

Finally, use the density of water to find its mass. The density of water is 62.3 lb/ft³. Therefore, \( 91.69 \text{ ft}^3 \times 62.3 \text{ lb/ft}^3 = 5,709.55 \text{ lb} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Unit Conversion
Unit conversion is a fundamental skill in mathematics and physics. It involves changing one unit of measure to another without altering the actual value. For example, when converting drops to gallons in the water calculation exercise, we used the given conversion factor: 46,000 drops are equal to one gallon. This allows us to transform a large number of drops into a smaller, more manageable number. To perform accurate conversions, multiply or divide by the appropriate factor. Remember: knowing your conversion factors is key!
Volume Measurement
In our daily lives and scientific calculations, measuring volume is essential. Volume is the amount of space an object or substance occupies. In our exercise, we measured the tap water leak in gallons per year. To illustrate, 46,000 drops of water amount to one gallon. However, to convert this volume to cubic feet, we used the factor that one gallon equals 0.133681 cubic feet. Using different units like liters, gallons, and cubic feet depends on your requirement but understanding how to switch between them is vital in solving problems effectively.
Mass Calculation
Knowing the mass of a substance helps us understand its quantity better. Mass calculation is often done by multiplying the volume by the density of the substance. In our exercise, we calculated the mass of the wasted water by first determining the volume in cubic feet and then using the density of water. With a volume of 91.69 cubic feet and water's density at 62.3 lb/ft³, the calculation becomes straightforward. Multiply these two values: 91.69 ft³ * 62.3 lb/ft³ = 5,709.55 lb. This calculation tells us how much the wasted water weighs, emphasizing the significance of understanding mass and volume relations.
Density of Water
Density is the mass per unit volume of a substance. For water, it's a crucial factor in many calculations. In our example, the density of water was given as 62.3 lb/ft³. This means that each cubic foot of water weighs 62.3 pounds. When we know the density, we can easily convert volume to mass. For instance, with the volume of wasted water calculated earlier as 91.69 cubic feet, we multiply this by the density to find the mass: 91.69 ft³ * 62.3 lb/ft³ = 5,709.55 lb. Understanding density helps in converting between mass and volume, especially in practical applications like understanding water wastage from a leaking tap.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Liquid water enters a valve at \(300 \mathrm{kPa}\) and exits at \(275 \mathrm{kPa}\). As water flows through the valve, the change in its temperature, stray heat transfer with the surroundings, and potential energy effects are negligible. Operation is at steady state. Modeling the water as incompressible with constant \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), determine the change in kinetic energy per unit mass of water flowing through the valve, in \(\mathrm{kJ} / \mathrm{kg}\).

Refrigerant \(134 \mathrm{a}\) flows at steady state through a horizontal tube having an inside diameter of \(0.05 \mathrm{~m}\). The refrigerant enters the tube with a quality of \(0.1\), temperature of \(36^{\circ} \mathrm{C}\), and velocity of \(10 \mathrm{~m} / \mathrm{s}\). The refrigerant exits the tube at 9 bar as a saturated liquid. Determine (a) the mass flow rate of the refrigerant, in \(\mathrm{kg} / \mathrm{s}\). (b) the velocity of the refrigerant at the exit, in \(\mathrm{m} / \mathrm{s}\). (c) the rate of heat transfer, in \(\mathrm{kW}\), and its associated direction with respect to the refrigerant.

Steam enters a turbine operating at steady state at \(700^{\circ} \mathrm{F}\) and \(450 \mathrm{lbf} / \mathrm{in}^{2}\) and leaves as a saturated vapor at \(1.2 \mathrm{lbf} / \mathrm{in} .^{2}\) The turbine develops \(12,000 \mathrm{hp}\), and heat transfer from the turbine to the surroundings occurs at a rate of \(2 \times 10^{6} \mathrm{Btu} / \mathrm{h}\). Neglecting kinetic and potential energy changes from inlet to exit, determine the volumetric flow rate of the steam at the inlet, in \(\mathrm{ft}^{3} / \mathrm{s}\).

A rigid tank having a volume of \(0.1 \mathrm{~m}^{3}\) initially contains water as a two-phase liquid-vapor mixture at 1 bar and a quality of \(1 \%\). The water is heated in two stages: Stage 1: Constant-volume heating until the pressure is 20 bar. Stage 2: Continued heating while saturated water vapor is slowly withdrawn from the tank at a constant pressure of 20 bar. Heating ceases when all the water remaining in the tank is saturated vapor at 20 bar. For the water, evaluate the heat transfer, in kJ, for each stage of heating. Ignore kinetic and potential energy effects.

Air enters a nozzle operating at steady state at \(720^{\circ} \mathrm{R}\) with negligible velocity and exits the nozzle at \(500^{\circ} \mathrm{R}\) with a velocity of \(1450 \mathrm{ft} / \mathrm{s}\). Assuming ideal gas behavior and neglecting potential energy effects, determine the heat transfer in Btu per lb of air flowing.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.