A hot 105.0 g lump of an unknown substance initially at 152.2 °C is placed in 35.0 mL of water initially at 25.0 °C and the system is allowed to reach thermal equilibrium. The final temperature of the system is 67.9 °C. Using this information and the specific heat values for several metals in the table, identify the unknown substance. Assume no heat is lost to the surroundings. tungsten graphite rhodium aluminum zinc titanium Substance Specific heat (J/(g.°C)) aluminum 0.897 graphite 0.709 rhodium 0.243 titanium 0.523 0.132 0.388 4.184 tungsten zinc water
A hot 105.0 g lump of an unknown substance initially at 152.2 °C is placed in 35.0 mL of water initially at 25.0 °C and the system is allowed to reach thermal equilibrium. The final temperature of the system is 67.9 °C. Using this information and the specific heat values for several metals in the table, identify the unknown substance. Assume no heat is lost to the surroundings. tungsten graphite rhodium aluminum zinc titanium Substance Specific heat (J/(g.°C)) aluminum 0.897 graphite 0.709 rhodium 0.243 titanium 0.523 0.132 0.388 4.184 tungsten zinc water
Chemistry
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ISBN:9781305957404
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Transcription and Analysis for Educational Purposes
**Problem Statement:**
A hot 105.0 g lump of an unknown substance initially at 152.2 °C is placed in 35.0 mL of water initially at 25.0 °C, and the system is allowed to reach thermal equilibrium. The final temperature of the system is 67.9 °C.
**Objective:**
Using this information and the specific heat values for several metals in the table, identify the unknown substance. Assume no heat is lost to the surroundings.
**Options:**
- Tungsten
- Graphite
- Rhodium
- Aluminum
- Zinc
- Titanium
**Table of Specific Heat Values:**
| Substance | Specific Heat (J/g·°C) |
|------------|------------------------|
| Aluminum | 0.897 |
| Graphite | 0.709 |
| Rhodium | 0.243 |
| Titanium | 0.523 |
| Tungsten | 0.132 |
| Zinc | 0.388 |
| Water | 4.184 |
**Approach to Solve the Problem:**
1. **Calculate the Heat Lost by the Unknown Substance:**
Use the formula:
\[
q = m \cdot c \cdot \Delta T
\]
where \( q \) is the heat exchanged, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Here, the specific heat is unknown, but the other variables can be used to set up the equation.
2. **Calculate the Heat Gained by the Water:**
Since no heat is lost to the surroundings, heat gained by water equals heat lost by the unknown substance. Use the same formula to calculate this with known specific heat for water.
3. **Set Heat Lost Equal to Heat Gained:**
\[
m_{\text{unknown}} \cdot c_{\text{unknown}} \cdot \Delta T_{\text{unknown}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}
\]
4. **Solve for the Specific Heat of the Unknown Substance:**
5. **Comparison:**
Compare the calculated specific heat with](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2659b4c-30a6-48d4-90ed-e5c7a02f7b8e%2F883dd424-7751-43a3-9cdc-d6ce19dcf5a9%2F51mh29a_processed.png&w=3840&q=75)
Transcribed Image Text:### Transcription and Analysis for Educational Purposes
**Problem Statement:**
A hot 105.0 g lump of an unknown substance initially at 152.2 °C is placed in 35.0 mL of water initially at 25.0 °C, and the system is allowed to reach thermal equilibrium. The final temperature of the system is 67.9 °C.
**Objective:**
Using this information and the specific heat values for several metals in the table, identify the unknown substance. Assume no heat is lost to the surroundings.
**Options:**
- Tungsten
- Graphite
- Rhodium
- Aluminum
- Zinc
- Titanium
**Table of Specific Heat Values:**
| Substance | Specific Heat (J/g·°C) |
|------------|------------------------|
| Aluminum | 0.897 |
| Graphite | 0.709 |
| Rhodium | 0.243 |
| Titanium | 0.523 |
| Tungsten | 0.132 |
| Zinc | 0.388 |
| Water | 4.184 |
**Approach to Solve the Problem:**
1. **Calculate the Heat Lost by the Unknown Substance:**
Use the formula:
\[
q = m \cdot c \cdot \Delta T
\]
where \( q \) is the heat exchanged, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Here, the specific heat is unknown, but the other variables can be used to set up the equation.
2. **Calculate the Heat Gained by the Water:**
Since no heat is lost to the surroundings, heat gained by water equals heat lost by the unknown substance. Use the same formula to calculate this with known specific heat for water.
3. **Set Heat Lost Equal to Heat Gained:**
\[
m_{\text{unknown}} \cdot c_{\text{unknown}} \cdot \Delta T_{\text{unknown}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}
\]
4. **Solve for the Specific Heat of the Unknown Substance:**
5. **Comparison:**
Compare the calculated specific heat with
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