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Use your data, the equation to the right, and the specific heat of water [tex]\(\left(4.184 \frac{J}{g^{\circ} C}\right)\)[/tex] to compute the specific heat values of each metal. Use a calculator and round to the nearest hundredth place.

Aluminum: [tex]\(c = \square \frac{J}{g^{\circ}C}\)[/tex]

Copper: [tex]\(c = \square \frac{J}{g^{\circ}C}\)[/tex]

Iron: [tex]\(c = \square \frac{J}{g^{\circ}C}\)[/tex]

Lead: [tex]\(c = \square \frac{J}{g^{\circ}C}\)[/tex]

\begin{tabular}{|c|c|c|c|c|}
\hline
& Al & Cu & Fe & Pb \\
\hline
[tex]$m_{\text{water}} (g)$[/tex] & 39.85 & 40.13 & 40.24 & 39.65 \\
\hline
[tex]$m_{\text{metal}} (g)$[/tex] & 11.98 & 12.14 & 12.31 & 12.46 \\
\hline
[tex]$\Delta T_{\text{water}} (^{\circ}C)$[/tex] & 4.7 & 1.9 & 2.4 & 0.7 \\
\hline
[tex]$\Delta T_{\text{metal}} (^{\circ}C)$[/tex] & -72.9 & -75.4 & -75.1 & -76.7 \\
\hline
\end{tabular}

Be sure to record all of these values in the data table.

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

Sagot :

GinnyAnsweridn
To determine the specific heat values of various metals using the given data, we can use the formula:

[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]

Given data:

- Specific heat of water, [tex]\( c_{\text{water}} = 4.184 \, \text{J/g°C} \)[/tex]
- Mass of water, [tex]\( m_{\text{water}} \)[/tex]:
- Aluminum: [tex]\( 39.85 \, \text{g} \)[/tex]
- Copper: [tex]\( 40.13 \, \text{g} \)[/tex]
- Iron: [tex]\( 40.24 \, \text{g} \)[/tex]
- Lead: [tex]\( 39.65 \, \text{g} \)[/tex]
- Mass of metal, [tex]\( m_{\text{metal}} \)[/tex]:
- Aluminum: [tex]\( 11.98 \, \text{g} \)[/tex]
- Copper: [tex]\( 12.14 \, \text{g} \)[/tex]
- Iron: [tex]\( 12.31 \, \text{g} \)[/tex]
- Lead: [tex]\( 12.46 \, \text{g} \)[/tex]
- Change in temperature of water, [tex]\( \Delta T_{\text{water}} \)[/tex]:
- Aluminum: [tex]\( 4.7 \, °\text{C} \)[/tex]
- Copper: [tex]\( 1.9 \, °\text{C} \)[/tex]
- Iron: [tex]\( 2.4 \, °\text{C} \)[/tex]
- Lead: [tex]\( 0.7 \, °\text{C} \)[/tex]
- Change in temperature of metal, [tex]\( \Delta T_{\text{metal}} \)[/tex]:
- Aluminum: [tex]\( -72.9 \, °\text{C} \)[/tex]
- Copper: [tex]\( -75.4 \, °\text{C} \)[/tex]
- Iron: [tex]\( -75.1 \, °\text{C} \)[/tex]
- Lead: [tex]\( -76.7 \, °\text{C} \)[/tex]

Let's compute the specific heat for each metal step-by-step:

1. Aluminum:
[tex]\[ c_{\text{Al}} = \frac{-4.184 \cdot 39.85 \cdot 4.7}{11.98 \cdot (-72.9)} \][/tex]
[tex]\[ c_{\text{Al}} = 0.90 \, \text{J/g°C} \][/tex]

2. Copper:
[tex]\[ c_{\text{Cu}} = \frac{-4.184 \cdot 40.13 \cdot 1.9}{12.14 \cdot (-75.4)} \][/tex]
[tex]\[ c_{\text{Cu}} = 0.35 \, \text{J/g°C} \][/tex]

3. Iron:
[tex]\[ c_{\text{Fe}} = \frac{-4.184 \cdot 40.24 \cdot 2.4}{12.31 \cdot (-75.1)} \][/tex]
[tex]\[ c_{\text{Fe}} = 0.44 \, \text{J/g°C} \][/tex]

4. Lead:
[tex]\[ c_{\text{Pb}} = \frac{-4.184 \cdot 39.65 \cdot 0.7}{12.46 \cdot (-76.7)} \][/tex]
[tex]\[ c_{\text{Pb}} = 0.12 \, \text{J/g°C} \][/tex]

So, the specific heat values for each metal rounded to the nearest hundredth are:
- Aluminum: [tex]\( c_{\text{Al}} = 0.90 \, \text{J/g°C} \)[/tex]
- Copper: [tex]\( c_{\text{Cu}} = 0.35 \, \text{J/g°C} \)[/tex]
- Iron: [tex]\( c_{\text{Fe}} = 0.44 \, \text{J/g°C} \)[/tex]
- Lead: [tex]\( c_{\text{Pb}} = 0.12 \, \text{J/g°C} \)[/tex]
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