Question

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

Answer 1

Answer:

65.47∘C

Explanation:

Specific heat capacity, c = 0.128 J/(g⋅∘C)

Initial temperature = 20.0 ∘C

Final temperature = ?

Mass = 52.4 g

Heat = 305 J

All these variables are related by the following equation;

H = m c ΔT

ΔT = H /  mc

ΔT = 305 / (52.4 * 0.128)

ΔT = 45.47∘C

ΔT = Final Temperature - Initial Temperature

Final temperature =  ΔT + Initial temperature

Final temperature = 45.47∘C + 20.0 ∘C = 65.47∘C