Question

Part E. Determination of a Calorimeter Constant A student doing a calorimetry experiment similar to this one is asked to find

Calculate Part E all the information is given

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Answer #1

For the first trial

calculate the heat gained by the cold water

Q cold = mass * specific heat * (Temperature difference)

Q cold = 50.1927 * 4.184 * (48.9 -19.7) =6132.18 Joules

calculate the heat lost by the hot water

mass of hot water = mass of total water - mass of cold water

mass of hot water = 75.22 - 50.197 = 25.0273 C

Q hot = 25.0273 * 4.184 * (48.9 - 83.5) = -3623.112 Joules

Now apply

Q calorimeter + Q hot + Q cool = 0

Q calorimeter = 3623.112 - 6132.18 = -2509.068 Joules

Q calorimeter = heat capacity * temperature difference

Temperature differece = (48.9 - 19.7) = 29.2 C

Q calorimeter = -2509.068 = heat capacity * 29.2

heat capacity = 2509.068 / 29.2 = 85.92 Joule / gram

Here is an image with the other calculations

75.22 74.3392 75.068 50.1927 50.1938 50.2011 25.0273 24.1454 24.8669 temp cold w final temp 19.7 48.9 20.1 49.5 19.9 52.2 tem

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