/*! 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 33 Nonmetric version: (a) How long ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Nonmetric version: (a) How long does a \(2.0 \times 10^{5}\) Btu/h water heater take to raise the temperature of 40 gal of water from \(70^{\circ} \mathrm{F}\) to \(100^{\circ} \mathrm{F}\) ? Metric version: (b) How long does a \(59 \mathrm{~kW}\) water heater take to raise the temperature of \(150 \mathrm{~L}\) of water from \(21^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
Both nonmetric and metric heaters take about 3 minutes to heat the water.

Step by step solution

01

Convert energy units for nonmetric version

The power of the heater is given in Btu/h, and we need to work with consistent units. Convert the energy being supplied by the heater from Btu/h to BTUs: 200,000 Btu/h heater delivers 200,000 Btu in one hour.
02

Calculate energy required to heat water (nonmetric)

The amount of energy required (Q) to heat the water can be calculated using the formula:\[ Q = m \cdot c \cdot \Delta T \]Where:- Q is the energy in Btu- m (mass) needs to be converted from gallons: 1 gallon of water is approximately 8.33 lbs - c is the specific heat of water (1 Btu/lb°F) - \Delta T\ is the temperature changeConvert gallons to pounds:\[ m = 40 \, \text{gal} \times 8.33 \, \text{lb/gal} = 333.2 \, \text{lb} \]Calculate temperature change:\[ \Delta T = 100^{\circ}F - 70^{\circ}F = 30^{\circ}F \]Thus:\[ Q = 333.2 \, \text{lb} \times 1 \, \text{Btu/lb°F} \times 30^{\circ}F = 9996 \, \text{Btu} \]
03

Calculate time required for nonmetric version

Using the calculated energy requirement, determine the time needed:\[ \text{Time (hours)} = \frac{Q}{\text{Power (Btu/h)}} = \frac{9996 \, \text{Btu}}{200,000 \, \text{Btu/h}} \approx 0.05 \, \text{hours} \approx 3 \, \text{minutes} \]
04

Convert energy units for metric version

The power of the heater is given in kW, and we need to work with consistent units. Convert the power to joules per second (J/s): 59 kW = 59,000 J/s.
05

Calculate energy required to heat water (metric)

The amount of energy required (Q) in joules to heat the water is calculated with:\[ Q = m \cdot c \cdot \Delta T \]Where:- Q is the energy in joules- m is the mass in kg (1 L of water has a mass of approximately 1 kg)- c is the specific heat of water (4200 J/kg°C)- \Delta T is the temperature changeCalculate mass:\[ m = 150 \, \text{kg (since 1 L = 1 kg)} \]Calculate temperature change:\[ \Delta T = 38^{\circ}C - 21^{\circ}C = 17^{\circ}C \]Thus:\[ Q = 150 \, \text{kg} \times 4200 \, \text{J/kg°C} \times 17^{\circ}C = 10,710,000 \, \text{J} \]
06

Calculate time required for metric version

Using the calculated energy requirement, determine the time needed:\[ \text{Time (seconds)} = \frac{Q}{\text{Power (J/s)}} = \frac{10,710,000 \, \text{J}}{59,000 \, \text{J/s}} \approx 181.53 \, \text{seconds} \approx 3.03 \, \text{minutes} \]

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.

Heat Transfer
Heat transfer is the process by which thermal energy is exchanged between physical systems. It's a fundamental concept in thermodynamics and is essential for understanding how energy moves. In this context, when heat is transferred into water, it causes the water's temperature to increase. This transfer occurs due to the water heater supplying energy into the water system.

There are three main methods of heat transfer: conduction, convection, and radiation. In the case of water heaters, conduction and convection are typically involved. Conduction occurs as heat moves through the heater's elements into the water. Convection happens as the heated water circulates and transfers heat throughout the tank.
  • Conduction: Direct transfer of heat through materials.
  • Convection: Transfer of heat by the movement of fluids.
By understanding which processes are at play, we can effectively calculate how long it takes for a given amount of heat to increase the temperature of water.
Specific Heat Capacity
Specific heat capacity is a property of a material that shows how much heat is needed to change its temperature. It is an essential factor in thermal calculations. For water, the specific heat capacity is relatively high, meaning it takes a lot of energy to change its temperature.

Using the formula \[ Q = m \cdot c \cdot \Delta T \] where:
  • \( Q \) is the amount of heat added (in joules or Btu).
  • \( m \) is the mass of the water (in kilograms or pounds).
  • \( c \) is the specific heat capacity \((1 \, ext{Btu/lb°F or 4200 \, J/kg°C})\).
  • \( \Delta T \) is the temperature change.
We can determine how much energy is necessary to heat the water. This concept helps students understand why substantial energy is required to raise water temperature, thus predicting and calculating heating times.
Energy Conversion
Energy conversion is about changing energy from one form to another, like converting electrical energy from a water heater into thermal energy in the water. This conversion is crucial for heating applications, ensuring the efficient performance of devices.

For instance, in the exercise provided:
  • In the nonmetric version, power is given in British Thermal Units (Btu), which is a unit of energy.
  • In the metric version, power is given in kilowatts (kW), a unit for measuring electricity.
The student must convert these units to a consistent energy form, allowing calculations that determine how long it will take to heat the water. Understanding conversions is vital as it forms the bridge between theoretical calculations and practical applications.
Temperature Change
Temperature change refers to the difference in temperature that occurs due to energy transfer into a substance. In heating water, knowing the initial and target temperatures is crucial to estimating the energy required.

In our calculation:
  • Nonmetric: The initial temperature is \( 70^{\circ}F \) raised to \( 100^{\circ}F \), yielding a temperature change \( \Delta T = 30^{\circ}F \).
  • Metric: The initial temperature is \( 21^{\circ}C \) raised to \( 38^{\circ}C \), yielding a temperature change \( \Delta T = 17^{\circ}C \).
Calculating \( \Delta T \) is pivotal since it directly influences how much energy is required. A larger temperature change generally means more energy. Understanding the relationship between temperature change and energy helps in designing efficient heating systems.

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

The area \(A\) of a rectangular plate is \(a b=1.4 \mathrm{~m}^{2}\). Its coefficient of linear expansion is \(\alpha=32 \times 10^{-6} / \mathrm{C}^{\circ} .\) After a temperature rise \(\Delta T=89^{\circ} \mathrm{C},\) side \(a\) is longer by \(\Delta a\) and side \(b\) is longer by \(\Delta b\) (Fig. 18-61). Neglecting the small quantity \((\Delta a \Delta b) / a b\), find \(\Delta A\).

A recruit can join the semi-secret "300 F" club at the Amundsen-Scott South Pole Station only when the outside temperature is below \(-70^{\circ} \mathrm{C}\). On such a day, the recruit first basks in a hot sauna and then runs outside wearing only shoes. (This is, of course, extremely dangerous, but the rite is effectively a protest against the constant danger of the cold. Assume that upon stepping out of the sauna, the recruit's skin temperature is \(102^{\circ} \mathrm{F}\) and the walls, ceiling, and floor of the sauna room have a temperature of \(30^{\circ} \mathrm{C}\). Estimate the recruit's surface area, and take the skin emissivity to be \(0.80 .\) (a) What is the approximate net rate \(P_{\text {net }}\) at which the recruit loses energy via thermal radiation exchanges with the room? Next, assume that when outdoors, half the recruit's surface area exchanges thermal radiation with the sky at a temperature of \(-25^{\circ} \mathrm{C}\) and the other half exchanges thermal radiation with the snow and ground at a temperature of \(-80^{\circ} \mathrm{C}\). What is the approximate net rate at which the recruit loses energy via thermal radiation exchanges with (b) the sky and (c) the snow and ground?

Consider the liquid in a barometer whose coefficient of volume expansion is \(6.6 \times 10^{-4} / \mathrm{C}^{\circ} .\) Find the relative change in the liquid's height if the temperature changes by \(12 \mathrm{C}^{\circ}\) while the pressure remains constant. Neglect the expansion of the glass tube.

A flow calorimeter is a device used to measure the specific heat of a liquid. Energy is added as heat at a known rate to a stream of the liquid as it passes through the calorimeter at a known rate. Measurement of the resulting temperature difference between the inflow and the outflow points of the liquid stream enables us to compute the specific heat of the liquid. Suppose a liquid of density \(0.85 \mathrm{~g} / \mathrm{cm}^{3}\) flows through a calorimeter at the rate of \(8.0 \mathrm{~cm}^{3} / \mathrm{s}\). When energy is added at the rate of \(250 \mathrm{~W}\) by means of an electric heating coil, a temperature difference of \(15 \mathrm{C}^{\circ}\) is established in steady-state conditions between the inflow and the outflow points. What is the specific heat of the liquid?

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of \(3.6 \mathrm{~kg}\) and contains \(14 \mathrm{~kg}\) of water. \(\mathrm{A} 1.8 \mathrm{~kg}\) piece of the metal initially at a temperature of \(180^{\circ} \mathrm{C}\) is dropped into the water. The container and water initially have a temperature of \(16.0^{\circ} \mathrm{C}\), and the final temperature of the entire (insulated) system is \(18.0^{\circ} \mathrm{C}\).

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.