486J of energy into 25mL of ethanol, how hot does it get if it were origin at 10.0 deg C?

Question

(C= 0.581 cal/g deg C; desity = 0.80g/mL)

its basically: (0.80g/mL)(0.581cal/g x degrees C)

... and I need help from here on.

[college chemisty 310 class; im a 10th grader in HS needing some guidance..1st college class.. thats all.]

Answer 1

1) Calculate the mass of ethanol
Mass = 25 mL x 0.8 g/mL = 20 g

2) Calculate the specific heat of ethanol.
1 cal = 4.18 J therefore
specific heat = 0.581 cal/(g deg C) * 4.18 J/cal = 2.43 J /(g deg C)

3) Calulate the temperature increase
temp increase = energy supplied / (mass * specific heat) = 486 J /(20 g * 2.43 J /(g deg C) ) = 10 deg C

Conclusion: the final temperature will be 20 deg C