Flag Thermal Physics> A 50 gm lead bullet specific heat 0.02 is...
question mark

A 50 gm lead bullet specific heat 0.02 is initially at 30 degree Celsius.It is fired vertically upwards with a speed of 840 m/s and on returning to the starting level, strikes a cake of ice at 0 degree celsius.Hw much ice is melted?? Latent heat of ice = 80 cal/g.Assume there is no loss in energy. Explain??

Navjyot Kalra , 11 Years ago
Grade 10
anser 1 Answers
Pragyan

Last Activity: 7 Years ago

Data:
Mass of lead bullet = m = 50 g = 50/1000 = 0.05 kg
Specific heat = 0.02 J
Initial temperature = 30°C
Speed = 840 m/s
Amount of melted ice = ?
Solution:
Total heat in a bullet = 50 x 0.02 x 30 
                                 = 30 cal
E = 1/2 mv² 
   = 1/2 (0.05) (840)²
   = 17640 J
Since, 1 J = 0.2389 cal
So, 17640 J = 4214.19 cal
Hence, E = 4214.19 cal 
Total heat = 4214.19 + 30 = 4244.19 J
Heat of fusion = 80 cal/g
So, total amount of melted ice = 4244.19 / 80 = 53 g
Hence, 53 g of ice is melted.
 
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments