a 100.00 mL sample of water has an initial temperature of 20.0 degrees C. The water is heated, and after 5 minutes its temperature is 50 degrees C. approximately how much energy was absorbed by the water? (the specific heat of wter is 4.18 J/g. degrees C

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  1. q = heat absorbed = mass H2O x specific heat H2O x delta T.

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  2. Q(H20)=100g x 4.18 JK-1g-1 x 30K
    =100 x 4.18J x 30 (Since the rest cancel each other out)

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