Question

A 5 kg hammer falls with a velocity 1 m/s on a piece of lead of mass 300g at 27° C. How many strokes will be required for melting the lead ?

Answer 1

M=5kg

v=1m/s

m=0.3kg

∆t=300°C

J=4.18×10^3J/Kcal

S=0.03Kcal/kg-°C

L=6 Kcal/Kg

Let no. of strokes will be n

From Joule's Law

W=J×Q

n× 1/2M×v^2=J(m×S×∆t + m×L)

n= 2×J×m(S×∆t+L)/M×v^2

n=2×4.18×10^3×0.3(0.03×300+6)/5×1^2

n=2508×15/5

n=7524