Asked by Joyce
. Determine the temperature, T that results when 150 g of ice at 0°C is mixed with 300 g of water at 50°C. Given heat of fusion, l of water = 3.35 × 105 J/kg, cwater = 4186 J/kg∙oC.
Answers
Answered by
bobpursley
Sum of heats gained is zero (some gains, some loses).
Heat ice + heat water=0
150*Hfice+150*cw*(Tf-0)+300*cw*(Tf-50)=0
solve for Tfinal, Tf
Heat ice + heat water=0
150*Hfice+150*cw*(Tf-0)+300*cw*(Tf-50)=0
solve for Tfinal, Tf
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