Answer to Question #331364 in General Chemistry for tim

Question

Compare the amount of heat given off by 1.40 mole of liquid water when it cools from 100.0 oC to 30.0 oC to

    that given off when 1.40 mol of steam cools from 200.0 ^oC to 110.0 ^oC   ( Cp H₂O _(l)  = 4.184 j/g-^oC  , Cp H₂O (g)

    = 1.87 J/j-^oC) . Explain your comparison

Answer 1

Mass of water = 1.40 * 18 = 25.2 (g)

Q₁ = cmΔT = 4.184 * 25.2 * (100.0 – 30.0) = 7380.6 J

Q₂ = 1.87 * 25.2 * (200.0 – 110.0) = 4241.2 J

Q₁ > Q₂