Question
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 $\mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$
Step 1
Substituting the given values, we get $q = 150.0 \, \text{g} \times 4.184 \, \text{J/g}^{\circ} \text{C} \times (28.0^{\circ} \text{C} - 24.6^{\circ} \text{C}) = 2131.8 \, \text{J}$. Show more…
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9.144 A 110.0 g piece of molybdenum metal is heated to 100.0 °C and placed in a calorimeter that contains 150.0 g of water at 24.6 °C. The system reaches equilibrium at a final temperature of 28.0 °C. Calculate the specific heat of molybdenum metal in J/(g • °C). The specific heat of water is 4.184 J/(g • °C).
A piece of metal weighing 58.657 g was heated to 97.0 °C and then placed into 150.0 g of water in a calorimeter at 20.5 °C. The metal and water were allowed to come to an equilibrium temperature, determined to be 26.5 °C. Assuming no heat lost to the environment, calculate the specific heat of the metal. The specific heat of water is 4.184.
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