Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 95

A British thermal unit (Btu) is defined as the quantity of heat required to change the temperature of 1 lb of water by \(1^{\circ}\) F. Assume the specific heat of water to be independent of temperature. How much heat is required to raise the temperature of the water in a 40 gal water heater from 48 to \(145^{\circ} \mathrm{F}\) in \((\mathrm{a}) \mathrm{Btu}\) (b) kcal; (c) kJ?

Short Answer

Expert verified
The required heat to raise the temperature of the water in a 40 gal water heater from 48 to \(145^{\circ} \mathrm{F}\) is \(32379.2 \, \mathrm{Btu}\), \(8163.5 \, \mathrm{kcal}\), or \(34139.8 \, \mathrm{kJ}\)

Step by step solution

01

Convert Volume to Weight

To find the weight of the water, multiply the volume by the density of the water. So, \(40 \, \mathrm{gal} \times 8.34 \, \mathrm{lb/gal} = 333.6 \, \mathrm{lb}\)
02

Find Temperature Difference

Subtract the initial temperature from the final temperature to find the difference. So, \(145^{\circ} \mathrm{F} - 48^{\circ} \mathrm{F} = 97^{\circ} \mathrm{F}\)
03

Calculate Heat in Btu

Insert the values into the heat formula \(Q = m \cdot c \cdot \Delta T\) where \(m = 333.6 \, \mathrm{lb}\), \(c = 1 \, \mathrm{Btu/lb^{\circ}F}\), and \(\Delta T = 97^{\circ} \mathrm{F}\). Therefore, \(Q = 333.6 \, \mathrm{lb} \times 1 \, \mathrm{Btu/lb^{\circ} F} \times 97^{\circ} \mathrm{F} = 32379.2 \, \mathrm{Btu}\).
04

Convert Btu to kcal and kJ

Use the conversion factors to convert Btu to kcal and kJ. So, for kcal: \(32379.2 \, \mathrm{Btu} \times 0.252 \, \mathrm{kcal/Btu} = 8163.5 \, \mathrm{kcal}\) and for kJ: \(32379.2 \, \mathrm{Btu} \times 1.055 \, \mathrm{kJ/Btu} = 34139.8 \, \mathrm{kJ}\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 internal energy of a fixed quantity of an ideal gas depends only on its temperature. A sample of an ideal gas is allowed to expand at a constant temperature (isothermal expansion). (a) Does the gas do work? (b) Does the gas exchange heat with its surroundings? (c) What happens to the temperature of the gas? (d) What is \(\Delta U\) for the gas?

A 75.0 g piece of \(\mathrm{Ag}\) metal is heated to \(80.0^{\circ} \mathrm{C}\) and dropped into \(50.0 \mathrm{g}\) of water at \(23.2^{\circ} \mathrm{C} .\) The final temperature of the \(\mathrm{Ag}-\mathrm{H}_{2} \mathrm{O}\) mixture is \(27.6^{\circ} \mathrm{C}\). What is the specific heat of silver?

Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic. $$\begin{array}{r} \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Fe}(\mathrm{s}) \\ \Delta H^{\circ}=-852 \mathrm{kJ} \end{array}$$ \(1.00 \mathrm{mol}\) of granular \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(2.00 \mathrm{mol}\) of granular Al are mixed at room temperature \(\left(25^{\circ} \mathrm{C}\right),\) and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat over a broad temperature range is about \(0.8 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) (The melting point of iron is \(1530^{\circ} \mathrm{C} .\) ) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the melting point of iron.

You are planning a lecture demonstration to illustrate an endothermic process. You want to lower the temperature of \(1400 \mathrm{mL}\) water in an insulated container from 25 to \(10^{\circ} \mathrm{C} .\) Approximately what mass of \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\) should you dissolve in the water to achieve this result? The heat of solution of \(\mathrm{NH}_{4} \mathrm{Cl}\) is \(+14.7 \mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{Cl}\).

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction \(\mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) given that $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+4 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow & 3 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ && \Delta H^{\circ}=-1937 \mathrm{kJ} \end{aligned}$$ $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.1 \mathrm{kJ} \end{array}$$
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation
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.

Sign-up for free