PRE-LAB Questions (show all the steps to calculations with correct significant figures and units for credit) A student mixed 50.0g of water at 80°C with 50.0 g of water at 20.0°C. The final temperature was 48.5°C. Calculate the heat capacity of the calorimeter.
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The formula for heat (q) is q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature. For water, the specific heat capacity (c) is 4.18 J/g°C. Show more…
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Lab Data Verify your calculation. Did you report your data to the correct number of significant figures? Temperature of cold water (°C) 2.0 Temperature of hot water (°C) 92.0 Volume of cold water (mL) 96.0 Volume of hot water (mL) 91.5 Final temperature after mixing (°C) 45.0 Mass of cold water (g) 96.0 Mass of hot water (g) 91.5 Calorimeter constant (J/°C) How to calculate the calorimeter constant
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A student obtains the following data in a calorimetry experiment designed to measure the specific heat of aluminum: $$\begin{array}{lc}\hline \begin{array}{l}\text { Initial temperature of water } \\\text { and calorimeter }\end{array} & 70^{\circ} \mathrm{C} \\ \text { Mass of water } & 0.400 \mathrm{~kg} \\\text { Mass of calorimeter } & 0.040 \mathrm{~kg} \\ \text { Specific heat of calorimeter } & 0.63 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C} \\\text { Initial temperature of aluminum } & 27^{\circ} \mathrm{C} \\\text { Mass of aluminum } & 0.200 \mathrm{~kg} \\ \text { Final temperature of mixture } & 66.3^{\circ} \mathrm{C} \\\hline\end{array}$$ Use these data to determine the specific heat of aluminum. Your result should be within $15 \%$ of the value listed in Table $20.1$.
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