Refer to this data table: Specific heat of: Molar heat of fusion of ice 6.01 kJ/mol. ice 2.03 J/g °C. Molar heat of vaporization of water = 40.79 kJ/mole steam = 1.99 J/g °C. water = 4.184 J/g °C. Calculate the heat in kJ theoretically required to transform 0.895 kg of ice at 0.00 °C to water at 38.0 °C. 92.3 kJ 263 kJ 822 kJ 318 kJ 194 kJ 710 kJ 74.6 kJ 441 kJ

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Chapter1: Chemical Foundations
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Refer to this data table:
Specific heat of:
Molar heat of fusion of ice = 6.01 kJ/mol.
ice = 2.03 J/g °C.
Molar heat of vaporization of water = 40.79 kJ/mole
steam = 1.99 J/g °C.
water = 4.184 J/g °C.
Calculate the heat in kJ theoretically required to transform 0.895 kg of ice at 0.00 °C to water at 38.0 °C.
92.3 kJ
263 kJ
822 kJ
318 kJ
194 kJ
710 kJ
74.6 kJ
441 kJ
Transcribed Image Text:Refer to this data table: Specific heat of: Molar heat of fusion of ice = 6.01 kJ/mol. ice = 2.03 J/g °C. Molar heat of vaporization of water = 40.79 kJ/mole steam = 1.99 J/g °C. water = 4.184 J/g °C. Calculate the heat in kJ theoretically required to transform 0.895 kg of ice at 0.00 °C to water at 38.0 °C. 92.3 kJ 263 kJ 822 kJ 318 kJ 194 kJ 710 kJ 74.6 kJ 441 kJ
Expert Solution
Step 1

Given,

Specific heat of

Ice = 2.03 J/g°C

Steam = 1.99 J/g°C

Water = 4.184 J/g°C

Molar heat of fusion of ice = 6.01 kJ/mol

Molar heat of vaporization of water = 40.79 kJ/mol

Mass of ice = 0.895 kg = 895 g

Initial temperature (Ti) = 0.00°C

Final temperature (Tf) = 38.0°C

Amount of heat required to transform 0.895 kg of ice at 0.00°C to water at 38.0°C = ?

 

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