Answer to Question #84878 in General Chemistry for klayne

Question #84878
What will be the final temperature of water that results when 5g of steam at 100°C is bubbled into 500g of water initially at 21°C
1
Expert's answer
2019-02-05T05:16:18-0500

Latent heat of vaporization ΔHvap = 2260 J/g

Specific heat of water c = 4.186 J/g*C

Q1 = Q2

Where Q1 is heat lost by steam

Q2 is heat gained by water

Q1 = m1ΔHvap + cm1 (T2 - T1)

Q2 = cm2(T2 - T1)

m1ΔHvap + cm1 (T2 - T1) = Q2 = cm2(T2 - T1)

As for the steam T2<T1 the value m1ΔHvap + cm1 (T2 - T1) will be negative, but as we should add m1ΔHvap + cm1 (T2 - T1), we should put "-" before m1ΔHvap + cm1 (T2 - T1)

m1ΔHvap - cm1 (T2 - T1) = Q2 = cm2(T2 - T1)

5g * 2260J/g - 4.186 J/g*C * 5 g *(T2 - 100C) = 4.186 J/g*C * 500 g *(T2 - 21C)

11300 - 20.93*T2 + 2093 = 2093*T2 - 43953

57346 = 2113.93* T2

T2 = 27C

Answer: 27 C

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