Physics B

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A 40-kg block of ice at 0°C is sliding on a horizontal surface. The initial speed of the ice is 7.0 m/s and the final speed is 4.7 m/s. Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice, and determine the mass of ice that melts into water at 0°C.

  • Physics B -

    Since the surface is horizontal, all of the loss of kinetic energy is due to frictional heating. The loss of kinetic energy is (1/2)M(Vfinal^2 - Vinitial^2)
    = 538 Joules = 129 calories

    If all that heat went into melting the ice, the amount that melted is 129 calories divided by the latent heat of fusion, which is 80 cal/gm

  • Physics B -

    The overall equation is Q=mL
    (m is mass; L is the latent heat of fusion for water)
    *we are trying to find mass (m)
    *rearranging the equation m=Q/L

    To find Q:
    the heat that caused the ice to melt was produced by the friction...we can find the amount of energy (in Joules) by finding KE (Kinetic Energy).

    *KE= .5 m (Vf^2-Vi^2)

    *when you find KE, use this value for Q and divide by the latent heat of fusion for water : 33.5x10^4 J/Kg

    Q(Joules)/L (Joules/Kg)=Kg

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