A scholar measures 20 mL of room temperature water (22ºC) and adds 80 mL of 75°C water to it, following the
procedure outlined on p. 2. The final temperature he reads is 80°C. Do you think this data is valid (yes or no?)
Support your answer using the first law of thermodynamics.

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

43.452

step-by-step explanation:

284 times .153 = 43.452

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being