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39 votes
39 votes
Calculate the change in heat (Q) that occurs when 7.58grams of steam at 101°C cools to liquid water at 40°C. (cp = 4.184 J/g°C)

User Ptah
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1 Answer

29 votes
29 votes

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

User Bethlee
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