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 37.7 g of this metal, initially at 20.0 ∘C?

Answer 1

q = mC∆T

q = heat = 305 J

m = mass = 37.7 g

C = specific heat = 0.128 J/gº

∆T = change in temperature = ?

Solving for ∆T:

∆T = q / (m)(C)

∆T = (305 J) / (37.7 g)(0.128 J/gº)

∆T = 63.2º

Since heat is being added to the metal, the temperature of the metal will rise by this amount.

Final temperature = 20.0º + 63.2º = 83.2ºC