andreaorthvolleyball
andreaorthvolleyball andreaorthvolleyball
  • 11-12-2016
  • Physics
contestada

15 g of lead (specific heat = 0.128 J/g Celsius) at 120 Degrees Celsius is placed on a glacier of ice at 0 degrees Celsius. How much ice melts? (Lf of ice = 333 J/g)

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MissPhiladelphia
MissPhiladelphia MissPhiladelphia
  • 12-12-2016

This can be solve using heat balance. The amount of heat the lead released is equal to the amount of heat the ice absorb.

The heat release by the leas = m1*Cp*deltaT

Where m1 is the mass of lead

Cp is the specific heat of lead

Delta is the change of temperature

Heat absorbed by the ice = m2Lf

Where m2 is the mass of ice

Lf is the latent heat of fusion of ice

 

M1*Cp*deltaT = m2Lf

(15g) ( 0.128 j/g C) ( 120 -0 C) = m2(333/g)

Solving for m2

M2 = 0.69 kg of ice melts

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