Question

beryllium is a rare metal that is gray in color, strong, and lightweight. the following data was obtained from a calorimeter experiment: - $m_{beryllium}$ = 10.1 g - $m_{water}$ = 38.8 g - $\delta t_{beryllium}$ = final $t$ - initial $t$ = 27.7°c - 100°c - $\delta t_{water}$ = final $t$ - initial $t$ = 28.4°c - 20.2°c - $c_{water}$ = 4.18 j/g°c the specific heat of beryllium is \boxed{} j/g°c. the formula given is $c_{metal} = \frac{-c_{water}m_{water}\delta t_{water}}{m_{metal}\delta t_{metal}}$

Explanation:

Step1: Calculate ΔT for beryllium and water

First, find \( \Delta T_{\text{beryllium}} \): \( \Delta T_{\text{beryllium}} = 27.7^\circ\text{C} - 100^\circ\text{C} = -72.3^\circ\text{C} \)
Then, find \( \Delta T_{\text{water}} \): \( \Delta T_{\text{water}} = 28.4^\circ\text{C} - 20.2^\circ\text{C} = 8.2^\circ\text{C} \)

Step2: Plug values into the formula

The formula for specific heat of the metal (\( c_{\text{metal}} \)) is:

$$ c_{\text{metal}} = \frac{-c_{\text{water}} m_{\text{water}} \Delta T_{\text{water}}}{m_{\text{metal}} \Delta T_{\text{metal}}} $$

We know:

• \( c_{\text{water}} = 4.18 \, \text{J/g}^\circ\text{C} \)
• \( m_{\text{water}} = 38.8 \, \text{g} \)
• \( \Delta T_{\text{water}} = 8.2^\circ\text{C} \)
• \( m_{\text{metal}} = 10.1 \, \text{g} \)
• \( \Delta T_{\text{metal}} = -72.3^\circ\text{C} \)

Substitute these values into the formula:

$$ c_{\text{metal}} = \frac{-(4.18 \, \text{J/g}^\circ\text{C})(38.8 \, \text{g})(8.2^\circ\text{C})}{(10.1 \, \text{g})(-72.3^\circ\text{C})} $$

Step3: Calculate the numerator and denominator

First, calculate the numerator:
\( -(4.18)(38.8)(8.2) \approx -(4.18 \times 38.8 \times 8.2) \approx -(1322.512) \)

Then, calculate the denominator:
\( (10.1)(-72.3) \approx -730.23 \)

Now, divide the numerator by the denominator:

$$ c_{\text{metal}} = \frac{-1322.512}{-730.23} \approx 1.81 \, \text{J/g}^\circ\text{C} $$

Answer:

The specific heat of beryllium is approximately \( \boldsymbol{1.81} \, \text{J/g}^\circ\text{C} \) (rounded to two decimal places, or depending on significant figures, it could be adjusted, but this is the calculated value).