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

\[\frac{ 125\cal }{ 60g*0.092\cal} = DeltaT = the new temperature \]

Answer 2

you're missing the initial temp.

q=m*C*\((T_f-T_i)\rightarrow\dfrac{q}{m*C}= T_f-T_i\rightarrow\dfrac{q}{m*C}\color{red}{+T_i}= T_f\)

Answer 3

Yeah Tfinal + Tinitial nice