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

Final T° is 45.5°C

Explanation:

Formula for calorimetry is:

Q = m . C .  ΔT , where

ΔT = Final T° - Initial T°

C = Specific heat

m = mass

Let's replace with the data given

305 J = 93.4 g . 0.128 J/g°C . (Final T° - 20°C)

305 J / 93.4 g . 0.128 J/g°C  = Final T° - 20°C

25.5°C = Final T° - 20°C → Final T° = 25.5°C + 20°C = 45.5°C

it is b that is the answer.