One liter of water at 56◦C is used to make iced
tea.
How much ice at 0 ◦C must be added to
lower the temperature of the tea to 20 ◦C?
The specific heat of water is 1 cal/g ·
◦ C and
latent heat of ice is 79.7 cal/g.
Answer in units of g.

22.6g

Explanation:

Mass of water(mw)=1liter=1000g

Final temperature=20°C

Temperature of ice=0°C

Temperature of water=56°C

Change in temperature of water=56-20=36

change in temperature of ice=20-0=20

Specific heat of water=1cal/g°C

Latent heat of ice=79.7cal/g

1000x1x36=mx79.7x20

36000=1594xm

Divide both sides by 1594

36000 ➗ 1594=1594xm ➗ 1594

22.6=m of ice

m of ice=22.6g

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!