Question

In: Chemistry

Suppose that 34.25 g of ice at -14.5

Expert Solution

The heat lost by water is equal to the heat gain by ice.

+Q ice = -Q water

Moles of ice = 34.25 / 18 = 1.9 mol

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So, let us calculate -Q ice first.

given, mass of ice = 34.25 g

Q1 = heat absorbed when 34.25 g of ice at -14.5 oC is converted to 0 oC ice

Q1 = m c dT

Q1 = 1.9 x 37.5 x 14.5 = 1033.125 J = 1.033 kJ

Q2 = heat absorbed when 34.25 g of ice at 0 oC is converted to 0 oC water

Q2 = m Hfus

Q2 = 1.9 x 6.01 = 11.42 kJ

Q3 = heat absorbed when 34.25 g of ice at 0 oC water is converted to x oC of water

Q3 = m c dT

Q3 = 1.9 x 75.28 x x

Q3 = 143.032x J = 0.143x kJ

Q ice = Q1 + Q2 + Q3

Q ice = 1.033 + 11.42 + 0.143x = 12.453 + 0.143x kJ

Q ice = 12.453 + 0.143x kJ

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Now, let us calculate -Q water

given, mass of water = 40.88 g

Moles of water = 40.88 / 18 = 2.27 mol

Q1 = heat released when 40.88g of water at 91.4 oC is converted to x oC.

Q1 = m c dT

Q1 = 2.27 x 75.28 x (x - 91.4) = 170.88x - 15618.94 J = 0.17088x - 15.618 kJ

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+Q ice = -Q water

12.453 + 0.143x = - [ 0.17088x - 15.618 ]

0.143x + 0.17088x = 15.618 - 12.453

0.31388x = 3.165

x = 10.08 oC = 283.08 K

queen_honey_blossom answered 2 months ago

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