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

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

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

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

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

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

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



Answer :

Alright, let's compute the specific heat values for each metal step-by-step using the given data and the equation:

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

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

Here are the given data:
- Mass of water, [tex]\( m_{\text{water}} \)[/tex]:
- Aluminum (Al): 39.85 g
- Copper (Cu): 40.13 g
- Iron (Fe): 40.24 g
- Lead (Pb): 39.65 g
- Mass of metal, [tex]\( m_{\text{metal}} \)[/tex]:
- Aluminum (Al): 11.98 g
- Copper (Cu): 12.14 g
- Iron (Fe): 12.31 g
- Lead (Pb): 12.46 g
- Change in temperature of water, [tex]\( \Delta T_{\text{water}} \)[/tex]:
- Aluminum (Al): 4.7 °C
- Copper (Cu): 1.9 °C
- Iron (Fe): 2.4 °C
- Lead (Pb): 0.7 °C
- Change in temperature of metal, [tex]\( \Delta T_{\text{metal}} \)[/tex]:
- Aluminum (Al): -72.9 °C
- Copper (Cu): -75.4 °C
- Iron (Fe): -75.1 °C
- Lead (Pb): -76.7 °C

Now, we will compute the specific heat values for each metal:

### For Aluminum (Al):
[tex]\[ c_{\text{Al}} = \frac{-4.184 \, \text{J/g}^\circ\text{C} \times 39.85 \, \text{g} \times 4.7 \, ^\circ\text{C}}{11.98 \, \text{g} \times (-72.9 \, ^\circ\text{C})} \][/tex]
[tex]\[ c_{\text{Al}} = \frac{-784.84948 \, \text{J}}{-873.882} \, \text{g}^\circ\text{C} \][/tex]
[tex]\[ c_{\text{Al}} \approx 0.90 \, \text{J/g}^\circ\text{C} \][/tex]

### For Copper (Cu):
[tex]\[ c_{\text{Cu}} = \frac{-4.184 \, \text{J/g}^\circ\text{C} \times 40.13 \, \text{g} \times 1.9 \, ^\circ\text{C}}{12.14 \, \text{g} \times (-75.4 \, ^\circ\text{C})} \][/tex]
[tex]\[ c_{\text{Cu}} = \frac{-318.42436 \, \text{J}}{-915.706} \, \text{g}^\circ\text{C} \][/tex]
[tex]\[ c_{\text{Cu}} \approx 0.35 \, \text{J/g}^\circ\text{C} \][/tex]

### For Iron (Fe):
[tex]\[ c_{\text{Fe}} = \frac{-4.184 \, \text{J/g}^\circ\text{C} \times 40.24 \, \text{g} \times 2.4 \, ^\circ\text{C}}{12.31 \, \text{g} \times (-75.1 \, ^\circ\text{C})} \][/tex]
[tex]\[ c_{\text{Fe}} = \frac{-403.69536 \, \text{J}}{-924.731} \, \text{g}^\circ\text{C} \][/tex]
[tex]\[ c_{\text{Fe}} \approx 0.44 \, \text{J/g}^\circ\text{C} \][/tex]

### For Lead (Pb):
[tex]\[ c_{\text{Pb}} = \frac{-4.184 \, \text{J/g}^\circ\text{C} \times 39.65 \, \text{g} \times 0.7 \, ^\circ\text{C}}{12.46 \, \text{g} \times (-76.7 \, ^\circ\text{C})} \][/tex]
[tex]\[ c_{\text{Pb}} = \frac{-116.4603 \, \text{J}}{-955.742} \, \text{g}^\circ\text{C} \][/tex]
[tex]\[ c_{\text{Pb}} \approx 0.12 \, \text{J/g}^\circ\text{C} \][/tex]

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