Answer:
Q = 19065.4 J
Step-by-step explanation:
Q₁ = the required energy for phase change of H₂O from vapor to liquid
Q₁ = mass of H₂O * heat of vaporization
Q₁ = 7.58 g * 2260 J/g
Q₁ = 17130.8 J
Q₂ = energy required to cool the water fro 101°C to 40°C
Q₂ = mass of H₂O * specific heat of H₂O * ΔT
Q₂ = 7.58 × (4.184 J/gC) × (101 - 40)
Q₂ = 1934.6 J
Q = Q₁ + Q₂
Q = (17130.8 + 1934.6) J
Q = 19065.4 J