7.3) A sample of metal block at 75.5 °C was placed in 250.0 g water with the temperature of 22.5 °C. When the two substances reached thermal equilibrium, the final temperature was 24.3 °C. Find the mass of the metal block. The specific heat capacity of the metal is 0.385 J/g.°C and the specific heat capacity of water is 4.184 J/g. °C. 150 g O 74 g O 120 g O 968

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
### Sample Problem: Thermal Equilibrium

#### Problem Statement:
A sample of a metal block at 75.5°C was placed in 250.0 g water with a temperature of 22.5°C. When the two substances reached thermal equilibrium, the final temperature was 24.3°C. Find the mass of the metal block. The specific heat capacity of the metal is 0.385 J/g°C and the specific heat capacity of water is 4.184 J/g°C.

##### Choices:
- [ ] 150 g
- [ ] 74 g
- [ ] 120 g
- [ ] 96 g

#### Explanation:
This problem involves the concept of thermal equilibrium, where two substances at different temperatures come into contact and exchange heat until they reach the same temperature.

#### Diagram or Graph (Not Provided in the Image):
If there were a diagram, it would likely depict:
1. **Initial Conditions:**
   - Metal Block: 75.5°C
   - Water: 22.5°C
2. **Final Condition:**
   - Both Metal Block and Water: 24.3°C

The heat lost by the metal equals the heat gained by the water, which can be expressed as:
\[ m_{metal} \cdot c_{metal} \cdot \Delta T_{metal} = m_{water} \cdot c_{water} \cdot \Delta T_{water} \]

Where:
- \( m \) is mass
- \( c \) is specific heat capacity
- \( \Delta T \) is the change in temperature (\(T_{initial} - T_{final}\))

#### Calculation Example:
- For the metal:
  \[ (m_{metal} \cdot 0.385 \, \text{J/g°C}) \cdot (75.5°C - 24.3°C) \]

- For the water:
  \[ (250.0 \, \text{g} \cdot 4.184 \, \text{J/g°C}) \cdot (24.3°C - 22.5°C) \]

Set the heat lost by the metal equal to the heat gained by the water and solve for \( m_{metal} \).

#### Solution:
To solve the problem step-by-step, please apply the heat exchange formula and solve for the mass of the metal block. The answer will be one
Transcribed Image Text:### Sample Problem: Thermal Equilibrium #### Problem Statement: A sample of a metal block at 75.5°C was placed in 250.0 g water with a temperature of 22.5°C. When the two substances reached thermal equilibrium, the final temperature was 24.3°C. Find the mass of the metal block. The specific heat capacity of the metal is 0.385 J/g°C and the specific heat capacity of water is 4.184 J/g°C. ##### Choices: - [ ] 150 g - [ ] 74 g - [ ] 120 g - [ ] 96 g #### Explanation: This problem involves the concept of thermal equilibrium, where two substances at different temperatures come into contact and exchange heat until they reach the same temperature. #### Diagram or Graph (Not Provided in the Image): If there were a diagram, it would likely depict: 1. **Initial Conditions:** - Metal Block: 75.5°C - Water: 22.5°C 2. **Final Condition:** - Both Metal Block and Water: 24.3°C The heat lost by the metal equals the heat gained by the water, which can be expressed as: \[ m_{metal} \cdot c_{metal} \cdot \Delta T_{metal} = m_{water} \cdot c_{water} \cdot \Delta T_{water} \] Where: - \( m \) is mass - \( c \) is specific heat capacity - \( \Delta T \) is the change in temperature (\(T_{initial} - T_{final}\)) #### Calculation Example: - For the metal: \[ (m_{metal} \cdot 0.385 \, \text{J/g°C}) \cdot (75.5°C - 24.3°C) \] - For the water: \[ (250.0 \, \text{g} \cdot 4.184 \, \text{J/g°C}) \cdot (24.3°C - 22.5°C) \] Set the heat lost by the metal equal to the heat gained by the water and solve for \( m_{metal} \). #### Solution: To solve the problem step-by-step, please apply the heat exchange formula and solve for the mass of the metal block. The answer will be one
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

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