current attempt in progress\na 45.4 - g sample of an unknown metal at 100.0°c is placed in a constant…

current attempt in progress\na 45.4 - g sample of an unknown metal at 100.0°c is placed in a constant - pressure calorimeter containing 43.4 g of water at 27.55°c. assume that the heat capacity of the calorimeter equals the heat capacity of the water it contains. the final temperature is 29.83°c. calculate the heat capacity of the metal and use the result to identify the metal.\no iridium (c = 0.131 j/g·k)\no copper (c = 0.385 j/g·k)\no beryllium (c = 1.82 j/g·k)\no lead (c = 0.130 j/g·k)\no nickel (c = 0.448 j/g·k)\no zinc (c = 0.389 j/g·k)\no platinum (c = 0.134 j/g·k)\no aluminum (c = 0.903 j/g·k)\no cadmium (c = 0.231 j/g·k)\no silver (c = 0.235 j/g·k)

current attempt in progress\na 45.4 - g sample of an unknown metal at 100.0°c is placed in a constant - pressure calorimeter containing 43.4 g of water at 27.55°c. assume that the heat capacity of the calorimeter equals the heat capacity of the water it contains. the final temperature is 29.83°c. calculate the heat capacity of the metal and use the result to identify the metal.\no iridium (c = 0.131 j/g·k)\no copper (c = 0.385 j/g·k)\no beryllium (c = 1.82 j/g·k)\no lead (c = 0.130 j/g·k)\no nickel (c = 0.448 j/g·k)\no zinc (c = 0.389 j/g·k)\no platinum (c = 0.134 j/g·k)\no aluminum (c = 0.903 j/g·k)\no cadmium (c = 0.231 j/g·k)\no silver (c = 0.235 j/g·k)

Answer

Explanation:

Step1: Calculate heat absorbed by water and calorimeter

The heat capacity of water $C_{water}=4.184\ J/g\cdot K$. Since the heat - capacity of the calorimeter equals the heat - capacity of the water it contains, the total heat capacity of the water - calorimeter system $C_{total}=2\times m_{water}\times C_{water}$, where $m_{water} = 43.4\ g$. The change in temperature of the water $\Delta T_{water}=T_f - T_{i}=29.83^{\circ}C - 27.55^{\circ}C=2.28^{\circ}C$. The heat absorbed by the water - calorimeter system $q_{abs}=C_{total}\Delta T_{water}=2\times43.4\ g\times4.184\ J/g\cdot K\times2.28\ K$. $q_{abs}=2\times43.4\times4.184\times2.28 = 2\times43.4\times9.54\ J=824.5\ J$.

Step2: Calculate heat released by the metal

The change in temperature of the metal $\Delta T_{metal}=T_{i}-T_{f}=100.0^{\circ}C - 29.83^{\circ}C = 70.17^{\circ}C$. Let the heat capacity of the metal be $C_{metal}$. The mass of the metal $m_{metal}=45.4\ g$. The heat released by the metal $q_{rel}=m_{metal}C_{metal}\Delta T_{metal}$. Since $q_{rel}=q_{abs}$ (assuming no heat loss to the surroundings), we have $45.4\ g\times C_{metal}\times70.17\ K = 824.5\ J$. $C_{metal}=\frac{824.5\ J}{45.4\ g\times70.17\ K}$. $C_{metal}=\frac{824.5}{45.4\times70.17}\ J/g\cdot K=\frac{824.5}{3185.7}\ J/g\cdot K\approx0.26\ J/g\cdot K$. The closest value to $0.26\ J/g\cdot K$ among the options is cadmium ($C = 0.231\ J/g\cdot K$).

Answer:

cadmium ($C = 0.231\ J/g\cdot K$)