The specific heat of a certain type of metal is 0.128 J/(g*^ * C) What is the final temperature if 305 of heat is added to 65.3g of this metal initially at 20.0 degrees * C

Answer 1

You didn't provide the units of the heat added. I will assume it is 305 JOULES. That being the case, we proceed as follows:

q = mC∆T

q = heat = 305 J

m = mass = 65.3 g

C = specific heat = 0.128 J/gº

∆T = change in temperature = ?

We will solve for ∆T:

∆T = q / (m)(C) = 305 J / (65.3 g)(0.128 J/gº)

∆T = 36.5º

Final temperature = 20.0º + 36.5º = 56.5ºC