A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 31.8 ◦C . In an attempt to cool the liquid, which has a ma

Question

A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 31.8 ◦C . In an attempt to cool the liquid, which has a mass of 161 g , 131 g of ice at 0.0 ◦C is added. At the time at which the temperature of the tea is 28.8 ◦C , find the mass of the remaining

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Mộc Miên 2 months 2021-07-30T00:46:49+00:00 1 Answers 2 views 0

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    2021-07-30T00:48:01+00:00

    Answer:

    Explanation:

    mass of liquid m₁ = 161 g

    temperature t₁ = 31.8

    final temperature t₂ = 28.8

    Let m g of ice melted to cool the liquid

    heat gained = mass x latent heat of fusion + mass x loss of temp x s heat of water

    = m x 80 + m x 1 x ( 31.8 – 28.8 ) ( latent heat of ice = 80 cals/g )

    = 83 m

    heat lost  = 161 x 1 x ( 31.8 – 28.8 ) ( specific heat of water = 1 cal / g / k )

    = 161 x 3

    heat lost = heat gained

    83 m = 161 x 3

    m = 5.82 g

    mass of remaining ice = 131 – 5.82

    = 125.18 g

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