A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0 °C to 35.5 °C. If the specific heat of aluminum is 0.900 A J/g°C, what is the mass of the sample (in grams)?

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
icon
Related questions
Question
### Determining the Mass of an Aluminum Sample Using Heat Absorption

**Problem Statement:**

A sample of aluminum absorbs \(50.1 \, \text{J}\) of heat, causing an increase in its temperature from \(20.0 \, ^\circ\text{C}\) to \(35.5 \, ^\circ\text{C}\). Given that the specific heat capacity of aluminum is \(0.900 \, \text{J/g}^\circ\text{C}\), calculate the mass of the aluminum sample in grams.

**Explanation and Solution:**

To find the mass of the aluminum sample, we can use the formula for heat transfer:

\[ q = m \cdot c \cdot \Delta T \]

Where:
- \( q \) is the heat absorbed or released (in Joules, \( \text{J} \)),
- \( m \) is the mass of the sample (in grams, \( \text{g} \)),
- \( c \) is the specific heat capacity (in \( \text{J/g}^\circ\text{C} \)),
- \( \Delta T \) is the change in temperature (in \( ^\circ\text{C} \)).

Rearranging to solve for \( m \):

\[ m = \frac{q}{c \cdot \Delta T} \]

Plugging in the given values:
- \( q = 50.1 \, \text{J} \),
- \( c = 0.900 \, \text{J/g}^\circ\text{C} \),
- \( \Delta T = 35.5^\circ\text{C} - 20.0^\circ\text{C} = 15.5^\circ\text{C} \),

\[ m = \frac{50.1 \, \text{J}}{0.900 \, \text{J/g}^\circ\text{C} \cdot 15.5^\circ\text{C}} \]

Calculate the mass:
\[ m = \frac{50.1}{0.900 \cdot 15.5} \]

\[ m = \frac{50.1}{13.95} \approx 3.59 \, \text{g} \]

Thus, the mass of the aluminum sample is approximately \(3.59
Transcribed Image Text:### Determining the Mass of an Aluminum Sample Using Heat Absorption **Problem Statement:** A sample of aluminum absorbs \(50.1 \, \text{J}\) of heat, causing an increase in its temperature from \(20.0 \, ^\circ\text{C}\) to \(35.5 \, ^\circ\text{C}\). Given that the specific heat capacity of aluminum is \(0.900 \, \text{J/g}^\circ\text{C}\), calculate the mass of the aluminum sample in grams. **Explanation and Solution:** To find the mass of the aluminum sample, we can use the formula for heat transfer: \[ q = m \cdot c \cdot \Delta T \] Where: - \( q \) is the heat absorbed or released (in Joules, \( \text{J} \)), - \( m \) is the mass of the sample (in grams, \( \text{g} \)), - \( c \) is the specific heat capacity (in \( \text{J/g}^\circ\text{C} \)), - \( \Delta T \) is the change in temperature (in \( ^\circ\text{C} \)). Rearranging to solve for \( m \): \[ m = \frac{q}{c \cdot \Delta T} \] Plugging in the given values: - \( q = 50.1 \, \text{J} \), - \( c = 0.900 \, \text{J/g}^\circ\text{C} \), - \( \Delta T = 35.5^\circ\text{C} - 20.0^\circ\text{C} = 15.5^\circ\text{C} \), \[ m = \frac{50.1 \, \text{J}}{0.900 \, \text{J/g}^\circ\text{C} \cdot 15.5^\circ\text{C}} \] Calculate the mass: \[ m = \frac{50.1}{0.900 \cdot 15.5} \] \[ m = \frac{50.1}{13.95} \approx 3.59 \, \text{g} \] Thus, the mass of the aluminum sample is approximately \(3.59
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 4 steps

Blurred answer
Knowledge Booster
Thermodynamics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY