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.

Question

Mathematics

Mohamed Frank

15 January, 21:25

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.

+5

Answers (1)

Know the Answer?

Gideon Cox · Answer 1

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.