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Today, 17:59

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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Answers (1)
  1. Today, 19:38
    0
    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
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