/*! 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 24 A certain substance has a mass p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 24

A certain substance has a mass per mole of \(50.0 \mathrm{~g} / \mathrm{mol}\). When \(314 \mathrm{~J}\) is added as heat to a \(30.0 \mathrm{~g}\) sample, the sample's temperature rises from \(25.0^{\circ} \mathrm{C}\) to \(45.0^{\circ} \mathrm{C}\). What are the (a) specific heat and (b) molar specific heat of this substance? (c) How many moles are in the sample?

Short Answer

Expert verified
(a) 0.523 J/g掳C, (b) 26.15 J/mol掳C, (c) 0.6 mol.

Step by step solution

01

Calculate the Change in Temperature

First, find the change in temperature of the sample. The initial temperature is \(25.0^{\circ}C\) and the final temperature is \(45.0^{\circ}C\). Subtract the initial temperature from the final temperature:\[ \Delta T = 45.0^{\circ}C - 25.0^{\circ}C = 20.0^{\circ}C \]
02

Use the Heat Formula to Find Specific Heat

The relationship between heat added, mass, specific heat, and change in temperature is given by the formula:\[ q = m \cdot c \cdot \Delta T \]Where:\( q = 314 \text{ J} \) (heat added), \( m = 30.0 \text{ g} \) (mass of the sample), \( \Delta T = 20.0^{\circ}C \) (temperature change).Rearrange the equation to solve for specific heat \( c \):\[ c = \frac{q}{m \cdot \Delta T} \]Substitute the values:\[ c = \frac{314 \text{ J}}{30.0 \text{ g} \cdot 20.0^{\circ}C} = \frac{314}{600} = 0.523 \frac{\text{J}}{\text{g} \cdot ^{\circ}C} \]
03

Calculate Moles of Substance

To calculate the number of moles in the sample, use the molar mass of the substance:\[ \text{Molar mass} = 50.0 \text{ g/mol} \]The number of moles \( n \) is given by:\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{30.0 \text{ g}}{50.0 \text{ g/mol}} = 0.6 \text{ mol} \]
04

Calculate Molar Specific Heat

The molar specific heat \( C_m \) is the specific heat in terms of moles. To find this, convert specific heat from per gram to per mole using the relationship:\[ C_m = c \times \text{molar mass} \]Given that \( c = 0.523 \frac{\text{J}}{\text{g} \cdot ^{\circ}C} \) and molar mass is \( 50.0 \text{ g/mol} \):\[ C_m = 0.523 \times 50.0 = 26.15 \frac{\text{J}}{\text{mol} \cdot ^{\circ}C} \]

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 91影视!

Key Concepts

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

Molar Specific Heat
Molar specific heat represents how much heat energy is required to raise the temperature of one mole of a substance by 1 degree Celsius. It's an important property that helps us understand how various substances handle energy.
To calculate the molar specific heat (\( C_m \), you need to first know the specific heat (\( c \)) of the substance. The specific heat is the heat required to raise the temperature of a specific mass (usually 1 gram) of a substance by 1 degree Celsius.
Once you have the specific heat, multiply it by the molar mass of the substance (given in grams per mole). This will convert the specific heat from 'per gram' to 'per mole', thus giving you the molar specific heat: \[ C_m = c \times \text{molar mass} \]
Using the values from the problem, the specific heat (\( c \)) is \( 0.523 \frac{\text{J}}{\text{g} \cdot ^{\circ}C} \) and the molar mass is 50.0 g/mol, resulting in a molar specific heat (\( C_m \)) of \( 26.15 \frac{\text{J}}{\text{mol} \cdot ^{\circ}C} \).
Understanding molar specific heat can help in real-world scenarios like designing heating systems or studying chemical reactions where precise temperature control is crucial.
Moles Calculation
Calculating the number of moles in a sample is fundamentally connected to the concept of molar mass. Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol).
To find the number of moles (\( n \)), you use the formula: \[ n = \frac{\text{mass}}{\text{molar mass}} \]
For example, if a substance has a mass of 30.0 grams and a molar mass of 50.0 g/mol (as in our exercise), you can substitute these values into the formula to find the moles: \[ n = \frac{30.0 \text{ g}}{50.0 \text{ g/mol}} = 0.6 \text{ mol} \]
This calculation shows us that the 30.0-gram sample contains 0.6 moles.
Understanding how to calculate moles is essential for converting between mass and moles in chemical equations. It's a skill you'll frequently use in chemistry when measuring substances and predicting outcomes of reactions.
Heat Transfer Formula
The heat transfer formula is a crucial tool in thermodynamics. It relates heat added, mass, specific heat, and temperature change of a substance. The equation is written as: \[ q = m \cdot c \cdot \Delta T \]
Where:
  • \( q \) is the heat added (in Joules),
  • \( m \) is the mass (in grams),
  • \( c \) is the specific heat (in Joules per gram per degree Celsius),
  • \( \Delta T \) is the change in temperature (in degrees Celsius).

To solve for the specific heat (\( c \)), you rearrange the formula: \[ c = \frac{q}{m \cdot \Delta T} \]
This equation helps determine how much energy a material can store or need to change its temperature. Using the given values:
  • \( q = 314 \text{ J} \),
  • \( m = 30.0 \text{ g} \),
  • \( \Delta T = 20.0^{\circ}C \),

Plug them into the rearranged formula to find specific heat: \[ c = \frac{314 \text{ J}}{30.0 \text{ g} \cdot 20.0^{\circ}C} = 0.523 \frac{\text{J}}{\text{g} \cdot ^{\circ}C} \]
Understanding this formula is pivotal for fields like engineering, meteorology, and even culinary arts. It helps predict and manage temperature changes in a wide variety of contexts.

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

Suppose \(200 \mathrm{~J}\) of work is done on a system and \(70.0\) cal is extracted from the system as heat. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of (a) \(W,\left(\right.\) b) \(Q\), and (c) \(\Delta E_{\text {int }}\) ?

Leidenfrost effect. \(\mathrm{A}\) water drop will last about \(1 \mathrm{~s}\) on a hot skillet with a temperature between \(100^{\circ} \mathrm{C}\) and about \(200^{\circ} \mathrm{C}\). However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance \(L\) in Fig. \(18-48)\). Let \(L=\) \(0.100 \mathrm{~mm}\), and assume that the drop is flat with height \(h=1.50 \mathrm{~mm}\) and bottom face area \(A=4.00 \times 10^{-6} \mathrm{~m}^{2}\). Also assume that the skillet has a constant temperature \(T_{s}=300^{\circ} \mathrm{C}\) and the drop has a temperature of \(100^{\circ} \mathrm{C}\). Water has density \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), and the supporting layer has thermal conductivity \(k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .\) (a) At what rate is energy conducted from the skillet to the drop through the drop's bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last?

On finding your stove out of order, you decide to boil the water for a cup of tea by shaking it in a thermos flask. Suppose that you use tap water at \(19^{\circ} \mathrm{C}\), the water falls \(32 \mathrm{~cm}\) each shake, and you make 27 shakes each minute. Neglecting any loss of thermal energy by the flask, how long (in minutes) must you shake the flask until the water reaches \(100^{\circ} \mathrm{C}\) ?

An object of mass \(6.00 \mathrm{~kg}\) falls through a height of \(50.0 \mathrm{~m}\) and, by means of a mechanical linkage, rotates a paddle wheel that stirs \(0.600 \mathrm{~kg}\) of water. Assume that the initial gravitational potential energy of the object is fully transferred to thermal energy of the water, which is initially at \(15.0^{\circ} \mathrm{C}\). What is the temperature rise of the water?

In a solar water heater, energy from the Sun is gathered by water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is \(20 \%\) (that is, \(80 \%\) of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of 200 \(\mathrm{L}\) of water in the tank from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) in \(1.0 \mathrm{~h}\) when the intensity of incident sunlight is \(700 \mathrm{~W} / \mathrm{m}^{2}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Electromagnetism

Read Explanation

Fundamentals of Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Dynamics

Read Explanation

Kinematics Physics

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.