A beaker of water sits in the sun until it reaches an equilibrium temperature of 30°CThe beaker is made of 100 g of aluminum and contains 200 g of waterIn an attempt to cool this system, 120 g of ice at 0°C is added to the water.(a) Determine how much ice remains after the system reaches equilibrium.answer in g .(b) Determine the final temperature of the system if instead 55 g of ice were originally added to the warm water.answer in °C .thanks for the help.please show the equations you used.
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a) To melt 120g of water would require QmLf 12080cal/g 9600 cal To cool the aluminum and water from 30 to 0 deg C would require (mcdelta T) 200g1cal/gC30 + 100g.22cal/gC30 6660calTherefore only enough ice would melt to remove 6660 cal from the cup and watertherefore Q 6660cal m80cal/g So 6660cal/80cal/g 83.25g of ice would melt leaving 120-83.25 36.8g of iceb) Now 55 g of ice would require 5580 4400cal to meltNow balancing the Q loos/gain yields 2001(30-Tf) + 100.22(30-Tf) 4400 +551(Tf-0) Solving for Tf yields (200+22+55)Tf 20030+2230 -4400 2260 277Tf So Tf 2260/277 8.16 degC
