Asked by Feather
Suppose that the 10.0 g of NH3 is formed in a bomb calorimeter, and that all of the evolved
heat for the exothermic reaction is completely absorbed by the surrounding 1000 g of water.
Find the change in temperature (ΔT), and the final temperature (TF) of the water if it is initially at 20.0 oC. Show all units.
You would use the formula q=(s)(m)(delta t)
(4.18 j (g x degree Celsius) ) (1000 g) (20.0 Celsius) = 8.36 x10^4
I got this answer do I rearrange the formula to get then Delta temperature q/(s)(m) = delta t ?
heat for the exothermic reaction is completely absorbed by the surrounding 1000 g of water.
Find the change in temperature (ΔT), and the final temperature (TF) of the water if it is initially at 20.0 oC. Show all units.
You would use the formula q=(s)(m)(delta t)
(4.18 j (g x degree Celsius) ) (1000 g) (20.0 Celsius) = 8.36 x10^4
I got this answer do I rearrange the formula to get then Delta temperature q/(s)(m) = delta t ?
Answers
Answered by
DrBob222
Your answer is hard to follow but I don't think it's right.
q = mass H2O x specific heat H2O x delta T
27,123 kJ = 1 kg x (4.184 kJ/kg*C) x delta T or if you prefer it can be
27,123 J = 1000g x 4.184 J/g*C x delta T.
If you want to do the last part first, which is what it appears you tried to do, it is. I have dT as 6.48 C.
q = mass x sp.h. x (Tf-Ti)
27,123 = 1000g x 4.184 J/g*C x (Tf - 20)
Solve for Tf and I get 26.48 which is too many significant figures but you can always round later.
q = mass H2O x specific heat H2O x delta T
27,123 kJ = 1 kg x (4.184 kJ/kg*C) x delta T or if you prefer it can be
27,123 J = 1000g x 4.184 J/g*C x delta T.
If you want to do the last part first, which is what it appears you tried to do, it is. I have dT as 6.48 C.
q = mass x sp.h. x (Tf-Ti)
27,123 = 1000g x 4.184 J/g*C x (Tf - 20)
Solve for Tf and I get 26.48 which is too many significant figures but you can always round later.
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