Asked by Meghan
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
Answered by
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
Answered by
Meghan
thank you!
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