0.595 g sample of steam at 107.4∘C is condensed into a container with 5.77g of water at 16.5∘C. What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18 J g⋅∘C, the specific heat of steam is 2.01 Jg⋅∘C, and Δ𝐻vap=40.7 kJ/mol.
Question
0.595 g sample of steam at 107.4∘C is condensed into a container with 5.77g of water at 16.5∘C. What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18 J g⋅∘C, the specific heat of steam is 2.01 Jg⋅∘C, and Δ𝐻vap=40.7 kJ/mol.
Solution
To solve this problem, we need to consider the energy changes that occur during the process.
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First, the steam will cool down to 100°C (the boiling point of water). The energy (q1) required for this can be calculated using the formula q = mcΔT, where m is the mass, c is the specific heat, and ΔT is the change in temperature.
q1 = (0.595 g)(2.01 J/g°C)(107.4°C - 100°C) = 8.89 J
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Next, the steam will condense into water at 100°C. The energy (q2) released during this process can be calculated using the formula q = mΔHvap, where ΔHvap is the heat of vaporization.
q2 = (0.595 g)(40.7 kJ/mol) = 24.22 kJ = 24220 J
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The energy released in steps 1 and 2 will be absorbed by the water, raising its temperature. The final temperature (Tf) can be calculated using the formula q = mcΔT, rearranged to ΔT = q/mc.
ΔT = (q1 + q2) / [(0.595 g + 5.77 g)(4.18 J/g°C)] = (8.89 J + 24220 J) / (6.365 g)(4.18 J/g°C) = 37.8°C
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The initial temperature of the water was 16.5°C, so the final temperature will be 16.5°C + 37.8°C = 54.3°C.
So, the final temperature of the water mixture is 54.3°C.
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