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
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.
2 answers
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
(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
1 answer