wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A lead ball at 30 C is dropped from a height of 6.2 km. The ball is heated due to the air resistance and it completely melts just before reaching the ground. The molten substance falls slowly on the ground. Calculate the latent heat of fusion of lead. Specific heat capacity of lead = 126 J kg1 C1 and melting point of lead = 330 C. Assume that any mechanical energy lost is used to heat the ball. Use g = 10 ms2.


A

2.4 × 104 cal/kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

2.4 × 105 cal/kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

2.4 × 104 J/kg

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

4.4 × 105 J/kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

2.4 × 104 J/kg


The initial potential energy of the ball,

Potential Energy = mgh

= m × (10 m s2) × (6.2 × 103 m)

= m × (6.2 × 104 m2 s2)
= m × (6.2 × 104 J kg1) -----(i)

Energy required to take the ball from 30 C to 330 C
= mL + mc(∆ T)
= m ( L + c(∆ T))
= m ( L + (126*300 ))
= m ( L + 37800) ---- (ii)

Since potential energy of the ball is converted to heat energy to melt the ball, equating (i) and (ii),


m(6.2 × 104) =m(L × 37800)

(6.2 × 104)37800 = L

L = 2.4 × 104 J kg1


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Calorimetry
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon