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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.

Respuesta :

MathPhys

Answer:

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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