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wolverine [178]
3 years ago
15

A scholar measures 20 mL of room temperature water (22ºC) and adds 80 mL of 75°C water to it, following the

Chemistry
1 answer:
Leni [432]3 years ago
4 0

Answer:

Data is not valid

Explanation:

When two liquids having different temperatures are mixed, regardless of the volumes, the final mix temperature will ALWAYS be between the initial temperature values.

1st Law Thermo => Law of Conservation of Energy => Energy can not be created nor destroyed, only changed in form. Mixing 22°C with 75°C will NOT result in a mix having a final temperature of 80°C.

∑ΔE = 0 => (mcΔT)₁ + (mcΔT)₂ = 0

[(20g)(1cal/g·°C)(Tₓ - 22°C)] + [(80g)(1cal/g·°C)(Tₓ - 75°C)] = 0

=> 20(Tₓ - 22) + 80(Tₓ - 75) = 0

=> 20Tₓ - 440 + 80Tₓ - 75 = 0

=> 100Tₓ = 440 + 75 = 515

=> Tₓ = (515/100)°C = 51.5°C final mix temperature

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<h3>Further explanation</h3>

The complete question

<em>A student is doing experiments with CO2(g). Originally, a sample of gas is in a rigid container at 299K and 0.70 atm. The student increases the temperature of the CO2(g) in the container to 425K.</em>

<em>Calculate the pressure of the CO₂ (g) in the container at 425 K.</em>

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Gay Lussac's Law

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