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.

in progress 0
RobertKer 3 days 2021-07-22T14:38:11+00:00 2 Answers 3 views 0

Answers ( )

    0
    2021-07-22T14:39:39+00:00

    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

    0
    2021-07-22T14:40:09+00:00

    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

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