Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 49

(a) What are the units of molar heat capacity? (b) What are the units of specific heat? (c) If you know the specific heat of copper, what additional information do you need to calculate the heat capacity of a particular piece of copper pipe?

Short Answer

Expert verified
(a) The units of molar heat capacity are joules per mole degree Celsius (or Kelvin), written as J/(mol·K). (b) The units of specific heat are joules per gram degree Celsius (or Kelvin), written as J/(g·K) or joules per kilogram degree Celsius (or Kelvin), written as J/(kg·K). (c) To calculate the heat capacity of a particular piece of copper pipe, we need the mass of the copper pipe and the specific heat of copper. Using the formula Q = mcΔT, we can then determine the heat capacity.

Step by step solution

01

(a) Units of Molar Heat Capacity

Molar heat capacity is defined as the amount of heat needed to raise the temperature of one mole of a substance by one degree Celsius (or one Kelvin). The units of molar heat capacity can be obtained by considering its definition. Heat is usually measured in joules (J), and the amount of substance is measured in moles (mol). Therefore, the units of molar heat capacity are joules per mole degree Celsius (or Kelvin), written as J/(mol·K).
02

(b) Units of Specific Heat

Specific heat is defined as the amount of heat needed to raise the temperature of one unit mass of a substance by one degree Celsius (or one Kelvin). Similar to molar heat capacity, the units of specific heat can be obtained by considering its definition. Heat is measured in joules (J), and the mass is measured in grams (g) or kilograms (kg). Therefore, the units of specific heat are joules per gram degree Celsius (or Kelvin), written as J/(g·K) or joules per kilogram degree Celsius (or Kelvin), written as J/(kg·K).
03

(c) Additional Information Required

To calculate the heat capacity of a particular piece of copper pipe, we need the following additional information: 1. The mass of the copper pipe: Mass is needed to calculate the heat capacity because heat capacity is related to the specific heat and mass. The heat capacity formula can be written as Q = mcΔT, where Q is the heat capacity, m is the mass, c is the specific heat, and ΔT is the change in temperature. 2. The specific heat of copper (which is given): This value is necessary to perform the calculation using the formula mentioned above. Once we have the mass of the copper pipe and the specific heat of copper, we can calculate the heat capacity of the copper pipe using the formula Q = mcΔT.

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

(a) What is meant by the term system in thermodynamics? (b) What is a closed system? (c) What do we call the part of the universe that is not part of the system?

Consider two solutions, the first being \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{CuSO}_{4}\) and the second \(50.0 \mathrm{~mL}\) of \(2.00 \mathrm{MKOH}\). When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(27.7^{\circ} \mathrm{C}\). (a) Before mixing, how many grams of Cu are present in the solution of \(\mathrm{CuSO}_{4} ?\) (b) Predict the identity of the precipitate in the reaction. (c) Write complete and net ionic equations for the reaction that occurs when the two solutions are mixed. (d) From the calorimetric data, calculate \(\Delta H\) for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 \(\mathrm{mL},\) and that the specific heat and density of the solution after mixing are the same as that of pure water.

(a) A serving of a particular ready-to-serve chicken noodle soup contains \(2.5 \mathrm{~g}\) fat, \(14 \mathrm{~g}\) carbohydrate, and \(7 \mathrm{~g}\) protein. Estimate the number of Calories in a serving. (b) According to its nutrition label, the same soup also contains \(690 \mathrm{mg}\) of sodium. Do you think the sodium contributes to the caloric content of the soup?

When a \(6.50-\mathrm{g}\) sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.18 ), the temperature rises from \(21.6^{\circ} \mathrm{C}\) to \(37.8^{\circ} \mathrm{C}\). Calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH})\) for the solution process $$ \mathrm{NaOH}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water.

Calculate the enthalpy change for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) $$ given the following enthalpies of reaction: $$ \begin{array}{ll} \mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{6}(s) & \Delta H=-1640.1 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & \Delta H=-2940.1 \mathrm{~kJ} \end{array} $$
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

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