## 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 te

Question

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.

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Physics
3 days
2021-07-22T14:38:11+00:00
2021-07-22T14:38:11+00:00 2 Answers
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## Answers ( )

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=mx79.7×20

36000=1594xm

Divide both sides by 1594

36000 ➗ 1594=1594xm ➗ 1594

22.6=m of ice

m of ice=22.6g

Answer:m= 361.0832 g

Explanation:This answer is identitical to the answer before me except you add together the latent heat of ice and the temperature change of ice instead of multiplyng them.

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=mx(79.7+20) <- only difference from the answer above

36000=99.7xm

Divide both sides by 99.7

22.6=m of ice

m of ice=361.0832g