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
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...
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![### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0d9066b-5fdf-42a6-b05e-754bdfae176c%2F5fe53813-88e0-48cd-b32e-eb7a6f04610d%2Fh0wsao_processed.jpeg&w=3840&q=75)
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
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