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

Answer 1

Q = mC_p(T₂ - T₁) where Q is the heat put in (in joules), "m" is the mass of the material (in grams), Cp is the Specific Heat (in J/(g °C)), T₁ is the initial temperature (at time "1") and T₂ is the temperature at time "2" (both in °C):

305 = (55.2)(0.128)(T₂ - 20)

305 = (7.0656)(T₂ - 20)

305/7.0656 = T₂ - 20

43.16689 = T₂ - 20

T₂ = 43.16689 + 20 = 63.16689

Rounding to 3 sig figs, T₂ (the final temperature) = 63.2 °C