Question

use your data, the equation to the right, 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.
aluminum:
$c = \square$ j/g°c
copper:
$c = \square$ j/g°c
iron:
$c = \square$ j/g°c
lead:
$c = \square$ j/g°c
be sure to record all of these values in the data table.
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$$\begin{array}{|c|c|c|c|c|} \\hline & \\text{al} & \\text{cu} & \\text{fe} & \\text{pb} \\\\ \\hline m_{\\text{water}} \\, (\\text{g}) & 39.85 & 40.13 & 40.24 & 39.65 \\\\ \\hline m_{\\text{metal}} \\, (\\text{g}) & 11.98 & 12.14 & 12.31 & 12.46 \\\\ \\hline \\delta t_{\\text{water}} \\, (\\circ \\text{c}) & 4.7 & 1.9 & 2.4 & 0.7 \\\\ \\hline \\delta t_{\\text{metal}} \\, (\\circ \\text{c}) & -72.9 & -75.4 & -75.1 & -76.7 \\\\ \\hline \\end{array}$$

$c_{\text{metal}} = \frac{-c_{\text{water}} m_{\text{water}} \delta t_{\text{water}}}{m_{\text{metal}} \delta t_{\text{metal}}}$

Explanation:

The formula for the specific heat of the metal is given by \( c_{\text{metal}} = -\frac{c_{\text{water}} m_{\text{water}} \Delta T_{\text{water}}}{m_{\text{metal}} \Delta T_{\text{metal}}} \), where \( c_{\text{water}} = 4.184 \, \text{J/g}^\circ\text{C} \). We will calculate \( c_{\text{metal}} \) for each metal (Al, Cu, Fe, Pb) using their respective \( m_{\text{water}} \), \( m_{\text{metal}} \), \( \Delta T_{\text{water}} \), and \( \Delta T_{\text{metal}} \) values from the table.

Aluminum (Al):

Step1: Identify values

\( m_{\text{water}} = 39.85 \, \text{g} \), \( m_{\text{metal}} = 11.98 \, \text{g} \), \( \Delta T_{\text{water}} = 4.7 \, ^\circ\text{C} \), \( \Delta T_{\text{metal}} = -72.9 \, ^\circ\text{C} \), \( c_{\text{water}} = 4.184 \, \text{J/g}^\circ\text{C} \)

Step2: Substitute into formula

\( c_{\text{Al}} = -\frac{(4.184)(39.85)(4.7)}{(11.98)(-72.9)} \)

Step3: Calculate numerator

\( (4.184)(39.85)(4.7) \approx 4.184 \times 39.85 \times 4.7 \approx 166.73 \times 4.7 \approx 783.63 \)

Step4: Calculate denominator

\( (11.98)(-72.9) \approx -873.34 \)

Step5: Compute \( c_{\text{Al}} \)

\( c_{\text{Al}} = -\frac{783.63}{-873.34} \approx \frac{783.63}{873.34} \approx 0.897 \approx 0.90 \, \text{J/g}^\circ\text{C} \) (Wait, but actual Al specific heat is ~0.90, but let's check calculation again. Wait, maybe miscalculation. Wait, \( \Delta T_{\text{water}} \) for Al is 4.7? Wait, no, looking at the table: the column for \( \Delta T_{\text{water}} \) (°C) has Al: 4.7, Cu:1.9, Fe:2.4, Pb:0.7. And \( \Delta T_{\text{metal}} \) (°C) for Al: -72.9, Cu:-75.4, Fe:-75.1, Pb:-76.7. Let's recalculate Al:

Numerator: \( 4.184 \times 39.85 \times 4.7 \)
\( 4.184 \times 39.85 \approx 4.184 \times 40 - 4.184 \times 0.15 \approx 167.36 - 0.6276 \approx 166.7324 \)
\( 166.7324 \times 4.7 \approx 166.7324 \times 4 + 166.7324 \times 0.7 \approx 666.9296 + 116.71268 \approx 783.64228 \)

Denominator: \( 11.98 \times (-72.9) = -11.98 \times 72.9 \approx -11.98 \times 70 - 11.98 \times 2.9 \approx -838.6 - 34.742 \approx -873.342 \)

So \( c_{\text{Al}} = - (783.64228) / (11.98 \times (-72.9)) = 783.64228 / 873.342 \approx 0.897 \approx 0.90 \, \text{J/g}^\circ\text{C} \) (matches expected ~0.90)

Copper (Cu):

Step1: Identify values

\( m_{\text{water}} = 40.13 \, \text{g} \), \( m_{\text{metal}} = 12.14 \, \text{g} \), \( \Delta T_{\text{water}} = 1.9 \, ^\circ\text{C} \), \( \Delta T_{\text{metal}} = -75.4 \, ^\circ\text{C} \)

Step2: Substitute into formula

\( c_{\text{Cu}} = -\frac{(4.184)(40.13)(1.9)}{(12.14)(-75.4)} \)

Step3: Calculate numerator

\( (4.184)(40.13)(1.9) \approx 4.184 \times 40.13 \times 1.9 \approx 167.90 \times 1.9 \approx 319.01 \)

Step4: Calculate denominator

\( (12.14)(-75.4) \approx -915.36 \)

Step5: Compute \( c_{\text{Cu}} \)

\( c_{\text{Cu}} = -\frac{319.01}{-915.36} \approx \frac{319.01}{915.36} \approx 0.348 \approx 0.35 \, \text{J/g}^\circ\text{C} \) (Wait, actual Cu specific heat is ~0.385, maybe miscalculation. Wait, recalculate numerator: 4.184*40.13=4.184*40 + 4.184*0.13=167.36 + 0.54392=167.90392; 167.90392*1.9=167.90392*2 - 167.90392*0.1=335.80784 - 16.790392=319.017448. Denominator: 12.14*75.4=12*75.4 + 0.14*75.4=904.8 + 10.556=915.356. So \( c_{\text{Cu}} = 319.017448 / 915.356 ≈ 0.3485 ≈ 0.35 \, \text{J/g}^\circ\text{C} \) (close to actual ~0.385, maybe table values have slight errors, but proceed with calculation)

Iron (Fe):

Step1: Identify values

\( m_{\text{water}} = 40.24 \, \text{g} \), \( m_{\text{metal}} = 12.31 \, \text{g} \), \( \Delta T_{\text{water}} = 2.4 \, ^\circ\text{C} \), \( \Delta T_{\text{metal}} = -75.1 \, ^\circ\text{C} \)

Step2: Substitute into formula

\( c_{\text{Fe}} = -\frac{(4.184)(40.24)(2.4)}{(12.31)(-75.1)} \)

Step3: Calculate numerator

\( (4.184)(40.24)(2.4) \approx 4.184 \times 40.24 \times 2.4 \approx 168.36 \times 2.4 \approx 404.06 \)

Step4: Calculate denominator

\( (12.31)(-75.1) \approx -924.48 \)

Step5: Compute \( c_{\text{Fe}} \)

\( c_{\text{Fe}} = -\frac{404.06}{-924.48} \approx \frac{404.06}{924.48} \approx 0.437 \approx 0.44 \, \text{J/g}^\circ\text{C} \) (Actual Fe specific heat ~0.449, close)

Lead (Pb):

Step1: Identify values

\( m_{\text{water}} = 39.65 \, \text{g} \), \( m_{\text{metal}} = 12.46 \, \text{g} \), \( \Delta T_{\text{water}} = 0.7 \, ^\circ\text{C} \), \( \Delta T_{\text{metal}} = -76.7 \, ^\circ\text{C} \)

Step2: Substitute into formula

\( c_{\text{Pb}} = -\frac{(4.184)(39.65)(0.7)}{(12.46)(-76.7)} \)

Step3: Calculate numerator

\( (4.184)(39.65)(0.7) \approx 4.184 \times 39.65 \times 0.7 \approx 165.90 \times 0.7 \approx 116.13 \)

Step4: Calculate denominator

\( (12.46)(-76.7) \approx -955.68 \)

Step5: Compute \( c_{\text{Pb}} \)

\( c_{\text{Pb}} = -\frac{116.13}{-955.68} \approx \frac{116.13}{955.68} \approx 0.1215 \approx 0.12 \, \text{J/g}^\circ\text{C} \) (Actual Pb specific heat ~0.13, close)

Final Calculations (Rounded to nearest hundredth):

- Aluminum: \( \boldsymbol{0.90} \, \text{J/g}^\circ\text{C} \)
- Copper: \( \boldsymbol{0.35} \, \text{J/g}^\circ\text{C} \)
- Iron: \( \boldsymbol{0.44} \, \text{J/g}^\circ\text{C} \)
- Lead: \( \boldsymbol{0.12} \, \text{J/g}^\circ\text{C} \)

(Note: These values are approximate based on the given table data and calculation. Slight discrepancies may exist due to rounding during steps, but the method follows the formula provided.)

Answers:

- Aluminum: \( \boxed{0.90} \, \text{J/g}^\circ\text{C} \)
- Copper: \( \boxed{0.35} \, \text{J/g}^\circ\text{C} \)
- Iron: \( \boxed{0.44} \, \text{J/g}^\circ\text{C} \)
- Lead: \( \boxed{0.12} \, \text{J/g}^\circ\text{C} \)

Answer:

s:

• Aluminum: \( \boxed{0.90} \, \text{J/g}^\circ\text{C} \)
• Copper: \( \boxed{0.35} \, \text{J/g}^\circ\text{C} \)
• Iron: \( \boxed{0.44} \, \text{J/g}^\circ\text{C} \)
• Lead: \( \boxed{0.12} \, \text{J/g}^\circ\text{C} \)