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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