Chemistry - Specific Heat Problems HELP PLS!!!!

Question

I need to now the answer and it has to include 3 sig figs.

1. What is the final temperature of 19.2 g of water if it absorbs 1820 J of heat and its inital temperature was 25.6ºC? 

Answer 1

Use the expression q = mC∆T

q = heat = 1820 J

m = mass = 19.2 g

C = specific heat of water = 4.184 J/g/deg

∆T = change in temperature = ?

Solve for ∆T:

1820 J = (19.2 g)(4.184 J/g/deg)(∆T)

∆T = 1820 J/(80.33 J/deg) = 22.66 degrees

Since the water ABSORBS the heat (as stated in the problem), the temperature of the water must INCREASE.

Thus FINAL TEMP = 25.6ºC + 22.66ºC = 48.26ºC = 48.3ºC (to 3 significant figures)