Question

a 52.0 g sample of an unknown metal at 99.0 °c was placed in a constant-pressure calorimeter containing 60.0 g of water at 24.0 °c. the final temperature of the system was found to be 28.4 °c. calculate the specific heat of the metal. (the heat capacity of water is 4.18 \\(\frac{j}{g \cdot ^\circ c}\\) and heat capacity of the calorimeter is 12.4 \\(\frac{j}{^\circ c}\\).) be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate heat gained by water

The formula for heat gained by water is \( q_{water} = m_{water} \times c_{water} \times \Delta T_{water} \). Here, \( m_{water} = 60.0 \, g \), \( c_{water} = 4.18 \, \frac{J}{g \cdot ^\circ C} \), and \( \Delta T_{water} = 28.4^\circ C - 24.0^\circ C = 4.4^\circ C \). So, \( q_{water} = 60.0 \times 4.18 \times 4.4 \). Calculating this: \( 60.0 \times 4.18 = 250.8 \), then \( 250.8 \times 4.4 = 1103.52 \, J \).

Step2: Calculate heat gained by calorimeter

The formula for heat gained by calorimeter is \( q_{calorimeter} = C_{calorimeter} \times \Delta T_{calorimeter} \). Here, \( C_{calorimeter} = 12.4 \, \frac{J}{^\circ C} \) and \( \Delta T_{calorimeter} = 28.4^\circ C - 24.0^\circ C = 4.4^\circ C \). So, \( q_{calorimeter} = 12.4 \times 4.4 = 54.56 \, J \).

Step3: Total heat gained by system (water + calorimeter)

Total heat gained \( q_{gained} = q_{water} + q_{calorimeter} = 1103.52 + 54.56 = 1158.08 \, J \). This heat is lost by the metal, so \( q_{metal} = -q_{gained} = -1158.08 \, J \) (negative because it's losing heat).

Step4: Calculate specific heat of metal

The formula for heat lost by metal is \( q_{metal} = m_{metal} \times c_{metal} \times \Delta T_{metal} \). Here, \( m_{metal} = 52.0 \, g \), \( \Delta T_{metal} = 28.4^\circ C - 99.0^\circ C = -70.6^\circ C \), and \( q_{metal} = -1158.08 \, J \). Rearranging for \( c_{metal} \): \( c_{metal} = \frac{q_{metal}}{m_{metal} \times \Delta T_{metal}} \). Plugging in values: \( c_{metal} = \frac{-1158.08}{52.0 \times (-70.6)} \). First, calculate denominator: \( 52.0 \times (-70.6) = -3671.2 \). Then, \( \frac{-1158.08}{-3671.2} \approx 0.315 \, \frac{J}{g \cdot ^\circ C} \).

Answer:

\( 0.315 \, \frac{J}{g \cdot ^\circ C} \)