Ask Question
26 February, 23:13

How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC?

+4
Answers (1)
  1. 26 February, 23:20
    0
    Answer: 1,013.32 cal * 4.18 J/cal = 4,235.68 J

    Explanation:

    1) dа ta:

    Water ⇒ C = 1 cal/g°C

    m = 65.8 g

    Ti = 31.5°C

    Tf = 36.9°C

    Heat, Q = ?

    2) Formula:

    Q = mCΔT

    3) Calculations:

    Q = 65.8g * 1 cal/g°C * (46.9°C - 31.5°C) = 1,013.2 cal

    4) You can convert from calories to Joules using the conversion factor:

    1 cal = 4.18 J

    ⇒ 1,013.32 cal * 4.18 J/cal = 4,235.68 J
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC? ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers