A block of ice of mass 50 kg is sliding on a horizontal plane. It starts with speed 5 ms-1 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 -1 , J = 4.2 J cal-1)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
2.86 g
b
3.86 g
c
0.86 g
d
1.86 g
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Given, m=50kg,v=5ms−1,L=80calg−1J=42Jcal−1 and m=?Heat lost, i.e. Q1 = work done = K (kinetic energy)=12mv2=12×50×52=625Jand heat gained, i.e. Q2=m'×L=80×m'×4.2∵ Q1 = Q2625=m'×8×42⇒m'=1.86g