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kherson [118]
3 years ago
11

A 2.7 kg block of ice at a temperature of 0.0◦C and an initial speed of 6.5 m/s slides across a level floor. If 3.3 × 105 J are

required to melt 1.0 kg of ice, how much ice melts, assuming that the initial kinetic energy of the ice block is entirely converted to the ice’s internal energy? Answer in units of kg.
Physics
1 answer:
bixtya [17]3 years ago
4 0

Answer:

The mass of ice melted is: 1.728*10^{-4}(Kg)

Explanation:

We need to remember that the latent heat of fusion is related as:L_{f} =\frac{Q}{m}, where Lf is the latent heat of fusion, Q is the amount of heat and m is the mass, and when there is a change of phase, e.g ice into liquid water, the temperature remains constant during this process. Now we calculate the kinetic energy of the block as: E_{k}=\frac{m*v^{2} }{2}=\frac{2.7*6.5^{2} }{2} =57.02(Joules), so this energy is used to change ice into liquid water, and knowing L_{f}=3.3*10^{5}(Joules/Kg), then we can replace in the equation the latent heat of fusion and get the mass of ice melted as:m=\frac{L_{f} }{Q} =\frac{57.04}{3.3*10^{5} } =1.728*10^{-4}(Kg)

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<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

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Height is the y-axis of the graph which is Force i.e. 30N

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According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

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\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}

\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}

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