Question

If 125 cal of heat is applied to a 60.0-g piece of copper at 21.0 ∘C , what will the final temperature be? The specific heat of copper is 0.0920 cal/(g⋅∘C)

Answer 1

q = mcΔt

where q is the heat (125 cal), m is the mass mass (60.0), c is the specific heat 0.0920 cal/(g⋅∘C), and Δt is the change in temperature. you can re-write Δt as (tf - ti) where tf is the final temp. and ti is the initial temperature (21 degrees C).

you can plug everything in and solve for tf