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 ?

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.

Yes, you are on the right track! To find the change in temperature (ΔT), you rearrange the equation q = (s)(m)(ΔT) to solve for ΔT.

So, the rearranged equation would be:
ΔT = q / (s)(m)

Using the values given in the problem:
- q is the heat absorbed by the water, which is 8.36 x 10^4 J (Joules)
- s is the specific heat capacity of water, which is 4.18 J/(g°C)
- m is the mass of water, which is 1000 g

Plugging in these values, you have:

ΔT = (8.36 x 10^4 J) / (4.18 J/(g°C) * 1000 g)

Now, you can simplify and solve for ΔT.