A block of ice of mass 50 kg is sliding on a horizontal plane. It starts with speed 5 m / s and stops after moving through some distance. The mass of ice that has melted due to friction between the block and the surface is (assuming that no energy is lost and latent heat of fusion of ice is 80 cal / g , J =4.2 J / cal )

Question

A block of ice of mass 50 kg is sliding on a horizontal plane. It starts with speed 5 m / s and stops after moving through some distance. The mass of ice that has melted due to friction between the block and the surface is (assuming that no energy is lost and latent heat of fusion of ice is 80 cal / g , J =4.2 J / cal ) (A) 2.86 g (B) 3.86 g (C) 0.86 g (D) 1.86 g

Tardigrade Admin · Accepted Answer

Given, m=50 kg , v =5 ms -1, L =80 cal / g J =4.2 J / cal Heat lost, i.e, Q1= work done = change in kinetic energy = K K = kinetic energy i.e., (1/2) mv 2=(1/2) × 50 × 52=625 J and heat gained, i.e. Q 2= m ' × L =80 × m ' × 4.2 heat lost = heat gained ∴ Q 1= Q 2 625= m ' × 8 × 42 m '=1.86 g

$2.86 g$

$3.86 g$

$0.86 g$

$1.86 g$

2.86g

3.86g

0.86g

1.86g

Given, m=50kg,v=5ms−1,L=80cal/g J=4.2J/cal Heat lost, i.e, Q1​= work done = change in kinetic energy =K K= kinetic energy i.e., 21​mv2=21​×50×52=625J and heat gained, i.e. Q2​=m′×L=80×m′×4.2 heat lost = heat gained ∴Q1​=Q2​ 625=m′×8×42 m′=1.86g

$2.86 g$

$3.86 g$

$0.86 g$

$1.86 g$