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Use your data, the given equation, and the specific heat of water (4.184 J/g°C) to compute the specific heat values of each metal. Use a calculator and round to the nearest hundredth place.

Given:
[tex]\[ c_{\text{metal}} = \frac{-c_{\text{water}} \cdot m_{\text{water}} \cdot \Delta T_{\text{water}}}{m_{\text{metal}} \cdot \Delta T_{\text{metal}}} \][/tex]

Data Table:

| | Al | Cu | Fe | Pb |
|------------|-------|-------|-------|-------|
| [tex]\( m_{\text{water}} (g) \)[/tex] | 39.85 | 40.13 | 40.24 | 39.65 |
| [tex]\( m_{\text{metal}} (g) \)[/tex] | 11.98 | 12.14 | 12.31 | 12.46 |
| [tex]\( \Delta T_{\text{water}} (°C) \)[/tex] | 4.7 | 1.9 | 2.4 | 0.7 |
| [tex]\( \Delta T_{\text{metal}} (°C) \)[/tex] | -72.9 | -75.4 | -75.1 | -76.7 |

Compute the specific heat values:

- Aluminum: [tex]\( c = 0.90 \, \text{J/g}°\text{C} \)[/tex]
- Copper: [tex]\( c = \square \, \text{J/g}°\text{C} \)[/tex]
- Iron: [tex]\( c = \square \, \text{J/g}°\text{C} \)[/tex]
- Lead: [tex]\( c = \square \, \text{J/g}°\text{C} \)[/tex]

Sagot :

GinnyAnsweridn
To compute the specific heat values for Copper (Cu), Iron (Fe), and Lead (Pb), we need to use the following given equation for each metal:

[tex]\[ c_{\text{metal}} = \frac{-c_{\text{water}} m_{\text{water}} \Delta T_{\text{water}}}{m_{\text{metal}} \Delta T_{\text{metal}}} \][/tex]

Given:
- The specific heat of water, [tex]\( c_{\text{water}} = 4.184 \text{ J/g} ^{\circ} \text{C} \)[/tex]

Let's calculate the specific heat values for each metal step-by-step.

### Copper (Cu):
1. Mass of water ([tex]\(m_{\text{water}}\)[/tex]): 40.13 g
2. Mass of metal ([tex]\(m_{\text{metal}}\)[/tex]): 12.14 g
3. Change in temperature of water ([tex]\(\Delta T_{\text{water}}\)[/tex]): 1.9 [tex]\(^{\circ}\)[/tex]C
4. Change in temperature of metal ([tex]\(\Delta T_{\text{metal}}\)[/tex]): -75.4 [tex]\(^{\circ}\)[/tex]C

Substitute these values into the equation:

[tex]\[ c_{\text{Cu}} = \frac{-4.184 \times 40.13 \times 1.9}{12.14 \times -75.4} \][/tex]

#### Simplification (performed by calculator):
[tex]\[ c_{\text{Cu}} \approx 0.35 \text{ J/g} ^{\circ} \text{C} \][/tex]

### Iron (Fe):
1. Mass of water ([tex]\(m_{\text{water}}\)[/tex]): 40.24 g
2. Mass of metal ([tex]\(m_{\text{metal}}\)[/tex]): 12.31 g
3. Change in temperature of water ([tex]\(\Delta T_{\text{water}}\)[/tex]): 2.4 [tex]\(^{\circ}\)[/tex]C
4. Change in temperature of metal ([tex]\(\Delta T_{\text{metal}}\)[/tex]): -75.1 [tex]\(^{\circ}\)[/tex]C

Substitute these values into the equation:

[tex]\[ c_{\text{Fe}} = \frac{-4.184 \times 40.24 \times 2.4}{12.31 \times -75.1} \][/tex]

#### Simplification (performed by calculator):
[tex]\[ c_{\text{Fe}} \approx 0.44 \text{ J/g} ^{\circ} \text{C} \][/tex]

### Lead (Pb):
1. Mass of water ([tex]\(m_{\text{water}}\)[/tex]): 39.65 g
2. Mass of metal ([tex]\(m_{\text{metal}}\)[/tex]): 12.46 g
3. Change in temperature of water ([tex]\(\Delta T_{\text{water}}\)[/tex]): 0.7 [tex]\(^{\circ}\)[/tex]C
4. Change in temperature of metal ([tex]\(\Delta T_{\text{metal}}\)[/tex]): -76.7 [tex]\(^{\circ}\)[/tex]C

Substitute these values into the equation:

[tex]\[ c_{\text{Pb}} = \frac{-4.184 \times 39.65 \times 0.7}{12.46 \times -76.7} \][/tex]

#### Simplification (performed by calculator):
[tex]\[ c_{\text{Pb}} \approx 0.12 \text{ J/g} ^{\circ} \text{C} \][/tex]

Therefore, the specific heat values are:

- Copper ([tex]\(Cu\)[/tex]): [tex]\( 0.35 \text{ J/g} ^{\circ} \text{C} \)[/tex]
- Iron ([tex]\(Fe\)[/tex]): [tex]\( 0.44 \text{ J/g} ^{\circ} \text{C} \)[/tex]
- Lead ([tex]\(Pb\)[/tex]): [tex]\( 0.12 \text{ J/g} ^{\circ} \text{C} \)[/tex]
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