30444 a.- 101lo40 2.) A coffee cup calorimeter contains 480.0 grams of water at
ID: 541726 • Letter: 3
Question
30444 a.- 101lo40 2.) A coffee cup calorimeter contains 480.0 grams of water at 25.00 °C. To it are added: 380.0 grams of water at 53.5 °O 525.0 grams of water at 65.5 °C Assuming the heat absorbed by the styrofoam is negligible, calculate the expected final temperature. The specific heat of water is 4.184 ig-1 -1 3.) A certain oil used in industrial transformers has a density of 1.086 g m1-1 and a specific heat of 1.826 J g-1 °C-1 . Calculate the heat capacity of one gallon of this oil. (1 gallon = 3.785 liters)Explanation / Answer
2. our governing equation will be H = m*s*T,
where H - heat, m - mass, s - specific heat, and T - temperature difference.
We will assumer our reference state to be at 25°C, to make calculations simpler.
heat content of water at 53.5°C = 380*4.184*(53.5-25) = 45312.72 J
heat content of water at 65.5°C = 525*4.184*(65.5-25) = 88962.3 J
total heat added = 45312.72+88962.3 = 134275.02 J
Rearranging the initial equation we get,
T = h/(m*s) = 134275.02/{(480+380+525)*4.184} = 23.17°C
Therefore the final temperature is 25+23.17 = 48.17°C
3. 1 gallon = 3.785L = 3785mL [since 1L = 1000mL]
total weight of the given oil = 1.086*3785 = 4110.51g [since density = mass/volume; mass = density*volume]
heat capacity = wt. * sp. heat = 4110.51*1.826 = 7505.79 J/°C
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