Question

1.
A piece of lead falls from a height of 100m
on a fixed non-conducting slab which brings
it to rest. If the specific heat of lead is
30.6 cal/kg °C, the increase in temperature
of the slab immediately after collision is
1) 6.72°C
2) 7.62°C
3) 5.62°C
4) 8.72°C​

Answer 1

Answer:

7.62 is the answer

Explanation:

Given,

h=100m

s=30.6cal/kg  

0

C

Potential energy of lead= thermal energy of slab

mgh=msΔT

ΔT=  

s

gh

​  

 

ΔT=  

30.6cal/kg  

0

C

10×100

​  

. . . . .(1)

we know that 1cal=4.2J

Equation (1) becomes,

ΔT=  

30.6×4.2

1000

​  

 

ΔT=7.62