Question
A hot stell ball of mass 1 kg is dropped into a 4 L bucket of water at room temperature (70 degrees F). If the system reaches an equilibrium temperature of 320 K, what was the initial temperature of the steel ball before it was dropped into the water? Cp of steel = 490 J/kg degrees C, Cp of water = 4186 J/kg degrees C.
Answers
bobpursley
the sum of heats gained is zero.
HeatgainedSteelBall+ HeatgainedWater=0
1kg*490*(Tf-Ti)+4000*4186*(Tf-70)=0
solve for Ti
HeatgainedSteelBall+ HeatgainedWater=0
1kg*490*(Tf-Ti)+4000*4186*(Tf-70)=0
solve for Ti
thank you!
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