Question
More than 150 years ago Pierre Dulong and A. T. Petit discovered a rule of thumb that the heat capacity of one mole of a pure solid element is about 6.0 calories per ${ }^{\circ} \mathrm{C}$ (i.e., about $25 \mathrm{~J} /{ }^{\circ} \mathrm{C}$ ). A $100.2-\mathrm{g}$ sample of an unknown metal at $99.9^{\circ} \mathrm{C}$ is placed in $50.6 \mathrm{~g}$ of water at $24.8^{\circ} \mathrm{C}$. The temperature is $36.6^{\circ} \mathrm{C}$ when the system comes to equilibrium. Assume all heat lost by the metal is absorbed by the water. What is the likely identity of this metal?
Step 1
The specific heat capacity of water is $4.18 \, \mathrm{J/g^{\circ}C}$, the mass of water is $50.6 \, \mathrm{g}$, and the change in temperature $\Delta T$ is $36.6^{\circ}C - 24.8^{\circ}C = 11.8^{\circ}C$. Show more…
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More than 150 years ago Pierre Dulong and A. T. Petit discovered a rule of tbumb that the heat capacity of one mole of a pure solid element is about $6.0$ calories per ${ }^{\circ} \mathrm{C}$ (i.e., about $\left.25 \mathrm{~J} /{ }^{\circ} \mathrm{C}\right)$. A $100.2-\mathrm{g}$ sample of an unknown metal at $99.9^{\circ} \mathrm{C}$ is placed in $50.6 \mathrm{~g}$ of water at $24.8^{\circ} \mathrm{C}$. The temperature is $36.6^{\circ} \mathrm{C}$ when the system comes to equilibrium. Assume all heat lost by the metal is absorbed by the water. What is the likely identity of this metal?
A 110.0 g piece of molybdenum metal is heated to $100.0^{\circ} \mathrm{C}$ and placed in a calorimeter that contains 150.0 $\mathrm{g}$ of water at $24.6^{\circ} \mathrm{C} .$ The system reaches equilibrium at a final temperature of $28.0^{\circ} \mathrm{C}$ . Calculate the specific heat of molybdenum metal in $\mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$ . The specific heat of water is 4.184 $\mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$
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