Question

How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 ∘C, to steam at 109.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

Answer 1

Mass of the Ice: 13 g

V1 = -10⁰C

V2 = 109⁰C

Cp (ice)= 2.09 J/g.⁰C

Cp (steam) = 2.01 J/g.⁰C

Answer:

q = m·ΔH_f

where
q = heat energy
m = mass
ΔH_f = heat of fusion

q for melting ice = (mass) (Δt) (C_{p of ice} )

q₁= 13 * (0-(-10)) * 2.09 J/g.⁰C = 271.7 J

For boiling water to vapour from 0⁰C to 109⁰C heat require is ,

q₂= 13 * [109-0] *2.01 = 2848.17 J

Since heat of heat of fusion of water and heat of vaporization of water is not mentioned in the question we will only add q₁ and q₂ to get total energy require to vaporize 13 g of ice

q_total = 271.7 J + 2848.17 J = 3119.87 J = 3.1198 KJ