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5 June, 15:40

Reaches an equilibrium temperature of 32°C.

In an attempt to cool the liquid, which has a

mass of 180 g, 112 g of ice at 0.0°C is added.

At the time at which the temperature of the

tea is 15C, find the mass of the remaining

ice in the jar. The specific heat of water

is 4186 J/kg. ° C. Assume the specific heat

capacity of the tea to be that of pure liquid

water.

Answer in units of g.

+2
Answers (1)
  1. 5 June, 18:57
    0
    77 g

    Step-by-step explanation:

    Heat gained by the ice = heat lost by the tea

    mL + mCΔT = - mCΔT

    m (334,000 J/kg) + m (2090 J/kg/°C) (15°C - 0°C) = - (0.180 kg) (4186 J/kg/°C) (15°C - 32°C)

    m (365,350 J/kg) = 12,809.16 J

    m = 0.035 kg

    m = 35 g

    35 grams of ice is melted and warmed to the final temperature, which leaves 77 grams unmelted.
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