CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A ball at 40C is dropped from a height of 5 km. The ball is heated due to air resistance and it completely melts just before reaching the ground. The specific heat capacity of the ball is 125 J kg1C1 and melting point is 340C. Then, the latent heat of the fusion of the ball (in kJ kg1) is

[Take g=10 m/s2]

Open in App
Solution

Let the mass of the ball be m kg.
Gravitational potential energy Q=mgh
Q=m×5×104 J
Energy required to heat the ball from 40C to 340C is given by
Q1=msΔT
Q1=m×125×[34040]=m×37500 J
Energy required to melt the ball completely
Q2=mL, where L = Latent heat of fusion of the ball

For the ball to melt completely, gravitational P.E must be converted to the required heat due to the presence of air resistance.
From the data given in the question,
Q=Q1+Q2
m×5×104=m×37500+mL
L=12500 J kg1
The latent heat of the fusion of ball is 12500 J kg1=12.5 kJ/kg

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon