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

Question

asked Jan 6, 2023 224k views

1 Answer

Asur · Answer 1 · 2023-01-13T00:48:25+0000

Answer:

6820 calories

Step-by-step explanation:

Specific heat of water = 1 c / gm-C

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