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.

Answer: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