Only 2, 3, 4 Thanks. A graph showing the idealized behavior of the system is sho
ID: 1009465 • Letter: O
Question
Only 2, 3, 4 Thanks.
A graph showing the idealized behavior of the system is shown below. The temperature axis is intentionally unlabeled and the data relevant to the determination of the calorimeter constant are in the table below the graph.
In the following questions, as in the laboratory, the heats and enthalpies in the exercise vary over a wide range. Almost all data are limited to 3 significant figures. To insure that your answers are properly interpreted, please give your answers in the units indicated in the questions but do not include the units with the answer you submit.
Question 1
What would the final temperature at the time of mixing, Tmix, have been if the calorimeter had been ideal ( Ccal = 0 )?
Enter Your Answer: 34.2C Correct
Question 2
How much heat (in kJ) was lost by the hot water? (Pay attention to the sign!)
Enter Your Answer: -3.42kJ Incorrect
Question 3
How much heat (in kJ) was gained by the cold water?
Enter Your Answer: 3.33kJ Incorrect
Question 4
What is the calorimeter constant (in J/oC)? (Note requested units!)
Enter Your Answer: 4.86J/oC Incorrect
Value Units Volume of cold water 50.8 mL Volume of hot water 47.9 mL Density of water (hot or cold) 1.00 g/mL Specific Heat of water (both hot and cold) 4.186 J/g-K Temperature of cold water at mixing, Tcold, (extrapolated from graph) 18.5 oC Temperature of hot water at mixing, Thot, (extrapolated from graph) 51.3 oC Temperature of mixture water at mixing, Tmix, (extrapolated from graph) 33.3 oCExplanation / Answer
final temperature at time of mixing = 34.2 C
Initial tempreature of cold water = 18.5
Initial temperature of hot water = 51.3
2) the heat lost by hot water = heat gained by cold water
heat lost by hot water = Mass of hot water X specific heat of water X change in temperture
Mass of water =Denisty x volume = 1 X 47.9 grams = 47.9 grams
Heat lost by hot water = 47.9 X 4.186 X (33.3-51.3) = 200.5094 X ( -18) = -3609.17 Joules = -3.609 KJ
3) heat gained by water = Mass of cold water X specific heat of water X change in temperature =
(50.8) X 4.186 X (33.3-18.5) = 3147.20 Joules = 3.147 KJ
4) the heat lost by hot water = Heat gained by cold water + heat gained by calorimeter
3.609 = 3.147 + heat gained by calorimeter
Heat gained by calorimeter = 0.462 KJ = 462 Joules
Change in temperature = 33.3-18.5 = 14.8
heat gained by calorimeter = Heat constant X change in temperature
462 = calorimeter constant X 14.8
Calorimeter constant = 462/14.8 = 31.22 J / C
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