204,024 views
26 votes
26 votes
How much heat in calories must be

added to change 100.0 grams of water
from 19.7 - 87 9°C?

User Dovholuk
by
3.3k points

1 Answer

13 votes
13 votes

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

User Steve Pike
by
2.6k points