A 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.

To solve this problem, we need to consider three steps:

1. The cooling of the steam from 107.4°C to 100°C.
2. The condensation of the steam at 100°C to water at 100°C.
3. The mixing of the hot water (from the condensed steam) with the cooler water until they reach a final equilibrium temperature.

Step 1: Cooling of the steam to 100°C The heat lost by the steam as it cools 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)(100°C - 107.4°C) = -14.9 J

Step 2: Condensation of the steam to water The heat released when the steam condenses can be calculated using the formula q = mΔHvap, where ΔHvap is the heat of vaporization.

First, we need to convert the mass of steam to moles, since ΔHvap is given per mole. The molar mass of water is approximately 18 g/mol, so we have 0.595 g / 18 g/mol = 0.033 mol.

q2 = (0.033 mol)(40.7 kJ/mol)(1000 J/1 kJ) = 1341 J

Step 3: Mixing of the hot water with the cooler water The heat gained by the cooler water as it warms up can be calculated using the formula q = mcΔT, where ΔT is the final temperature minus the initial temperature.

Let's call the final temperature T. Then we have:

q3 = (5.77 g)(4.18 J/g°C)(T - 16.5°C)

Since no heat is lost, the sum of the heats in all three steps should be zero:

q1 + q2 + q3 = 0

Substituting the expressions for q1, q2, and q3 gives:

-14.9 J + 1341 J + (5.77 g)(4.18 J/g°C)(T - 16.5°C) = 0

Solving this equation for T gives the final temperature of the water mixture.