Question

Consider a glass of 169 mL of water at 29

Answer 1

Q ice = -Q water

Let us find Q ice.

Q1 = heat absorbed when -15 oC is converted to 0oC ice

Q1 = m c delta T

Q1 = m x 2.01 J/g K x 15

Q1 = 30.15 m J

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Q2 : heat absorbed when 0 oC ice is converted to 0oC water

Q2 = m x heat of fusion

Q2 = m x 334 J/g

Q2 = 334 m J

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Q3 : heat absorbed when 0 oC water is converted to 10 oC water

Q3 = m c delta T

Q3 = m x 4.18 J/g K x 10

Q3 = 41.8 m J

Q ice = Q1 + Q2 + Q3

Q ice = 30.15 m J + 334 m J + 41.8 m J

Q ice = 405.95 m J

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Now, let us find Q water

Mass of water = volume of water x density = 169 ml x 1g / ml = 169 g

Q1 = heat released when 169 g of water at 29 oC is converted to 10 oC water

Q1 = m c delta T

Q1 = 169 x 4.18 x (-19)

Q1 = -13421.98 J

Thus, Q water = -13421.98 J

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Q ice = -Q water

405.95 m J = -(-13421.98 J)

m = 33.1 g

Thus, 33.1 g of ice must be added.