20.0 g of ice cubes at 0.0°C are combined with 150. g of liquid water at 70.0°C in a coffee cup calorimeter. Calculate the final temperature reached, assuming no heat loss or gain from the surroundings. (Data: specific heat capacity of H20( ), c = 4.18 J/g•°C; H 20( s) – H20() A H = 6.02 kJ/mol)Hint: When you equate heat lost to heat gained the heat capacity is cancelled since it si found on both sides of the equation and this simplifies your calculation.

Chemistry by OpenStax (2015-05-04)
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Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
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Chapter5: Thermochemistry
Section: Chapter Questions
Problem 27E: The addition of 3.15 g of Ba(OH)28H2O to a solution of 1.52 g of NH4SCN in loo g of water in a...
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20.0 g of ice cubes at 0.0°C are combined with 150. g of liquid water at 70.0°C in a coffee cup calorimeter. Calculate the final
temperature reached, assuming no heat loss or gain from the surroundings. (Data: specific heat capacity of H20(), c = 4.18
J/g.°C; H 20( s) – H20() A H= 6.02 kJ/mol)Hint: When you equate heat lost to heat gained the heat capacity is
cancelled since it si found on both sides of the equation and this simplifies your calculation.
O 0.0
O 10.6
O 30.7
O 56.4
O61.8
Transcribed Image Text:20.0 g of ice cubes at 0.0°C are combined with 150. g of liquid water at 70.0°C in a coffee cup calorimeter. Calculate the final temperature reached, assuming no heat loss or gain from the surroundings. (Data: specific heat capacity of H20(), c = 4.18 J/g.°C; H 20( s) – H20() A H= 6.02 kJ/mol)Hint: When you equate heat lost to heat gained the heat capacity is cancelled since it si found on both sides of the equation and this simplifies your calculation. O 0.0 O 10.6 O 30.7 O 56.4 O61.8
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