Chemistry

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13 Dec 2019

100. g of ice at 0Â°C is added to 300.0 g of water at 60Â°C. Assuming no transfer of heat to the surroundings, what is the temperature of the liquid water after all the ice has melted and equilibrium is reached?

specific heat (ice) = 2.10 J/gÂ°C

specific heat (water) = 4.18 J/gÂ°C

heat of fusion = 333 J/g

heat of vaporization = 2258 J/g

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Determine

how much

energy in kJ would be transferred if 300.0 grams

of steam (gaseous

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is cooled to ice

(solid water)

at -20.0

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The specific

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1.86

J/(gâ ÌC),

the

specific heat

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the heat of fusion of water

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the heat

of vaporization is 2256 J/g.

A 100.0 g sample of water at 80Â°C, is mixed with 10 g of ice cubesat 0Â°C. The heat of fusion of ice is 335 J/g, and the specific heatof liquid water is 4.18 J/gÂ°C. Calculate the final temperature indegrees Celsius of the mixture.

A. 0.0
B. 10.6
C. 30.7
D. 65
E. 53.2

burgundygrasshopper8

1a.Calculate the enthalpy change (in kJ) on converting 62.2 gramsof ice at 0Â°C to water at 0Â°C. The heat of fusion of water is 6.01kJ/mol

b.Calculate the heat required (in kJ) to convert 94.8 grams of iceat -25Â°C to water at 50Â°C.

The specific heats of ice and water are 2.09 J/gÂ°C and 4.18 J/gÂ°C,respectively, and the heat of fusion of ice is 6.01 kJ/mol.

plumdolphin370

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