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We are going to determine the specific heat of a metal using experimental data. In this experiment, we used a "coffee cup" calorimeter and gathered the following data:
Mass of empty cup 2.31 g Mass of cup + water 180.89 g Mass of cup + water + metal 780.89 g Initial temperature of water 17.0 °C Initial temperature of metal 52.0 °C Final temperature of system 27.0 °C
The key thermochemistry equation for solving this problem is:
qmetal = qwater
Then, by substitution, we have (metal values on the left, water values on the right):
(mass) (Δt) (Cp) = (mass) (Δt) (Cp)
We need to work with values from the data table to get what we need to substitute into the above equation.
mass of water: 180.98 - 2.31 = 178.58 g
mass of metal: 780.89 - 180.89 = 600.0 g
change in water temperature: 27.0 - 17.0 = 10.0 °C
change in metal temperature: 52.0 - 17.0 = 25.0 °C
Putting the numbers in place gives us:
(600.0 g) (25.0 °C) (x) = (178.58 g) (10.0 °C) (4.184 J g¯1 °C¯1)
Solving gives 0.498 J g¯1 °C¯1
Notice the starting temperature of the metal (52.0 °C). This is an unusual value in that the metal sample is usually heated up by immersion in boiling water, making the usual starting temperature at or near 100.0 °C for the metal.
Often, problems of this sort will specify mL of water, rather than grams. The mass of water present is determined by volume times density. Since the density of water is 1.00 g mL¯1, the calculation is:
your mL x 1.00 g mL¯1
with the answer being the same numerical value, just with grams as the unit rather than mL.