Answered

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How much heat in calories must be
added to change 100.0 grams of water
from 19.7 - 87 9°C?

Sagot :

Answer:

6820 calories

Explanation:

Specific heat of water =  1 c / gm-C

100 gm * (87.9-19.7) C *  1 c/gm-C = 6820 cal

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