how many joules are needed to change 120.g of water at 20. 0 c to steam at 100. 0 c? how many joules...

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any joules are needed to change 120.g at 20. 0 C to steam at 100. 0 C? ny joules of energy must be removed 26 g of water at 24 0 C to form ice at 0. many joules are required to change 225 ce at -15. o C to steam at 140. o C ?

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Page 1: How many joules are needed to change 120.g of water at 20. 0 C to steam at 100. 0 C? How many joules of energy must be removed from 126 g of water at 24

How many joules are needed to change 120.g ofwater at 20.0C to steam at 100.0C?

How many joules of energy must be removed from 126 g of water at 240C to form ice at 0.0C?

How many joules are required to change 225 g of ice at -15. oC to steam at 140. oC ?

Page 2: How many joules are needed to change 120.g of water at 20. 0 C to steam at 100. 0 C? How many joules of energy must be removed from 126 g of water at 24

How many joules are needed to change 120.g ofwater at 20.0C to steam at 100.0C?

First heat H2O from 20.0C to boiling (1000C):

Heat = 120.g x 80.0C x 4.18 J/g0C = 40128 J = 40.1 kJ

Then find heat to vaporize the water:

Heat = 120.g x 2260 J/g = 271200 J = 271 kJ Total heat = 311 kJ

Page 3: How many joules are needed to change 120.g of water at 20. 0 C to steam at 100. 0 C? How many joules of energy must be removed from 126 g of water at 24

How many joules of energy must be removed from 126 g of water at 240C to form ice at 00C?

First cool water from 240C to 00C:

Heat = 126g x 240C x 4.18 J/g0C = 12640 J (12.6 kJ) = 13 kJ

Then find heat removed to freeze H2O:

Heat = 126 g x 335 J/g = 42210 J = 42.2 kJ Total heat = 55 kJ

(54.8)

Page 4: How many joules are needed to change 120.g of water at 20. 0 C to steam at 100. 0 C? How many joules of energy must be removed from 126 g of water at 24

How many joules are required to change 225 g of ice at -15. oC to steam at 140. oC?

First heat the solid (-15. oC) to melting point(00C) : Heat = 225g x 15.oC x 2.03 J/g oC = 6851 J

= 6.9 kJ Then melt the ice:Heat = 225 g x 335 J/g = 75375 J = 75.4 kJ

Then heat the ice water(0.0C) to boiling(100.0C):Heat = 225 g x 100.0C x 4.18 J/g0C = 94050 J = 94.1 kJ

Then, find heat to vaporize the H2O:Heat = 225 g x 2259 J/g = 508275 J = (508.3) = 508 kJ

Total heat = 702 kJ Heat = 225g x 40.oC x 2.01 J/g oC =18090 J = 18 kJ

Finally, heat the gas (100. oC) to 1400C) :