Ask Question
27 February, 16:15

Steam (water vapor) at 100 degrees Celsius is added to a thermally insulated container with 200 kg of ice at zero degrees Celsius. The final mixture is water at 30 degrees Celsius. What was the initial mass of the steam?

+2
Answers (1)
  1. 27 February, 20:13
    0
    m = 359.24 kg

    Explanation:

    given dа ta:

    mass of ice = 200 kg

    latent heat of steam = 2260 kJ/kg

    latent heat of ice = 334 kJ/kg

    from conservatioon of energy principle we know that

    heat lost by steam will be equal to heat gained by ice

    therefore we have

    (ml + mcdt) = (ml + mcdt)

    m*2.26*10^6 + m*4186 * (100-30) = 200*3.33 * 10^6 + 2000*4186 * 30

    m = 359.24 kg
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Steam (water vapor) at 100 degrees Celsius is added to a thermally insulated container with 200 kg of ice at zero degrees Celsius. The ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers