Heat and phase change: latent heat

AI Thread Summary
The discussion centers on calculating the mass of ice that melts due to kinetic friction as a block of ice slides on a surface. The initial and final speeds of the ice are provided, and the heat generated by friction is assumed to completely melt a small portion of the ice. Participants express confusion over the equation used, particularly regarding the definition of Q and its dimensional consistency. They suggest that the kinetic energy lost should equal the heat gained by the ice, leading to the equation for mass melted based on latent heat of fusion. The conversation emphasizes the need for clarity in the equations used for energy and heat transfer calculations.
alaa410
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Homework Statement



A 42kg block of ice at 0°C is sliding on a horizontal surface. the initial speed of the ice is 7.3 m/s and the final speed is 3.5m/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. Determine the mass of the ice the melts into water

Homework Equations



Q=mL

The Attempt at a Solution



Q=mL + (1/2v^2 final- 1/2v^2initial)?
 
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alaa410 said:
Q=mL + (1/2v^2 final- 1/2v^2initial)?
What exactly is Q in that equation?
The right hand side is dimensionally inconsistent. It mixes energy and velocity2.
 
q is heat gained or lost; this is the equation they used in the solution so its got to be right...
 
except they multiply not add
 
alaa410 said:
q is heat gained or lost; this is the equation they used in the solution so its got to be right... except they multiply not add
Not exact enough for Q, and the equation would still be nonsensical.
I suggest that the equations ought to say
KE lost = heat gained by ice = mass melted * latent heat of fusion
So now you just need an expression for the KE lost.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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