Question

a 32.0 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 j/(g·°c) and heat capacity of the calorimeter is 11.4 j/°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, $q_{water} = 60.0\ g \times 4.18\ \frac{J}{g\cdot^\circ C} \times 4.4^\circ C$.
Calculating this: $60.0\times4.18\times4.4 = 60.0\times18.392 = 1103.52\ J$.

Step2: Calculate heat gained by calorimeter

The formula for heat gained by calorimeter is $q_{calorimeter} = C_{calorimeter} \times \Delta T_{calorimeter}$. Here, $C_{calorimeter} = 11.4\ \frac{J}{^\circ C}$ and $\Delta T_{calorimeter} = 28.4^\circ C - 24.0^\circ C = 4.4^\circ C$.
So, $q_{calorimeter} = 11.4\ \frac{J}{^\circ C} \times 4.4^\circ C = 49.76\ J$.

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

$q_{gained} = q_{water} + q_{calorimeter} = 1103.52\ J + 49.76\ J = 1153.28\ J$.

Step4: Heat lost by metal

By the principle of calorimetry, heat lost by metal ($q_{metal}$) is equal to heat gained by the system. So, $q_{metal} = -q_{gained} = -1153.28\ J$ (negative sign indicates heat loss).

Step5: 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, $m_{metal} = 32.0\ g$, $\Delta T_{metal} = 28.4^\circ C - 99.0^\circ C = -70.6^\circ C$. We need to find $c_{metal}$.
Rearranging the formula: $c_{metal} = \frac{q_{metal}}{m_{metal} \times \Delta T_{metal}}$.
Substituting the values: $c_{metal} = \frac{-1153.28\ J}{32.0\ g \times (-70.6^\circ C)}$.
First, calculate the denominator: $32.0\times(-70.6) = -2259.2\ g\cdot^\circ C$.
Then, $c_{metal} = \frac{-1153.28}{-2259.2} \approx 0.510\ \frac{J}{g\cdot^\circ C}$.

Answer:

$0.510\ \frac{J}{g\cdot^\circ C}$