The specific heat of a certain type of metal is 0.128 J/(g-c). What is the final temperature if 305 J of heat is added to 86.9 g of this metal initially at 20 degrees C? Tfinal = ___ C

Answer 1

q = 305J = mass metal x specific heat x (Tfinal-Tinitial). Substitute and solve for Tfinal.

Answer 2

ohhhhhhhh so 305J = (86.9)(0.128)(Tf-20)
305J = (11.1232)(Tf-20C)
305J = 11.1232[Tf] - 222.464
527.464 = 11.1232[Tf]
Tf = 47.42 C

is that correct..?