Asked by Jacob
The heat of reaction for burning 1 mole of a certain compound X is known to be -477.7 kJ. The calorimeter constant of the bomb being used is 2500 j/degree celsius and the initial temperature of the water is 23.2 degrees celsius.
a) if 22.54 g of compound X (MM=46.0) is burned in the bomb calorimeter containing 2000. ml of water (S.H.=4.184 j/g(degrees celsius), what will be the final temperature?
b) How much water can be warmed from 23.2 C to 56.5 C when 52.3 g of the compound is burned in the bomb?
a) if 22.54 g of compound X (MM=46.0) is burned in the bomb calorimeter containing 2000. ml of water (S.H.=4.184 j/g(degrees celsius), what will be the final temperature?
b) How much water can be warmed from 23.2 C to 56.5 C when 52.3 g of the compound is burned in the bomb?
Answers
Answered by
DrBob222
22.54/46.0 = moles cmpd X.
heat generated by 22.54 moles = 477.7 kJ/mole x (22.54/46.0) = ?? kJ.
Then heat generated (change to Joules) = mass water x specific heat water x (Tfinal-Tinitial) + 2500 J/C*(Tfinal-Tinitial)
Solve for Tfinal.
The second problem is the same concept.
heat generated by 22.54 moles = 477.7 kJ/mole x (22.54/46.0) = ?? kJ.
Then heat generated (change to Joules) = mass water x specific heat water x (Tfinal-Tinitial) + 2500 J/C*(Tfinal-Tinitial)
Solve for Tfinal.
The second problem is the same concept.
Answered by
Matthew
i still keep messing up, can someone give me a specific answer i can use as a base?
Answered by
DrBob222
I estimated the answer to the first problem as about 45 C or so.
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