Physics: Heat - Google Chrome Secure https://session.masteringphysics.com/myct/i
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Physics: Heat - Google Chrome Secure https://session.masteringphysics.com/myct/itemView?assignmentproblemID=91907625 Heat Adding Ice to Water 80114 ) Constants I Periodic Table Part A An insulated beaker with negligible mass contains liquid water with a mass of 0.260 kg and a temperature of 67.0°( How much ice at a temperature of -100 °C must be dropped into the water so that the final temperature of the system will be 40.0 °C? Take the specific heat of liquid water to be 4190 J/kg. K, the specific heat of ice to be 2100 J/kg- K and the heat of fusion for water to be 3.34x105 J/kg View Available Hint(s) kg mice Submit Next > Provide FeedbackExplanation / Answer
Mass of water M = 0.260 kg
Initial temperature of water t = 60.0o C
Mass of ice m = m
Initial temperature of ice t ' = -10o C
The final temperature of the system T =40.0o C
Specific Heat of liquid water C = 4190 J/kg/oC
Specific Heat of ice c =2100 J/kg/oC
Latent Heat of Fusion of water L = 3.34x 10^5 J/kg
Heat lost by water = heat gain by ice
MC(t-T) = mc(0-t ') + mL + mC(T-0)
21788 = m[21000 + 334000 +167600 ]
From this mass of ice m = 0.04169 kg
= 41.69 g