Question

1. A small bag containing 0.200 kg of lead shot at a temperature of 15.0C falls from a 40.0 m high tower. Instead of bouncing back, the bag makes a small hole in the ground. The specific heat of lead is 1.28  102 Jkg•C. a. Find the initial potential energy of the lead. b. How much energy did the lead lose as heat? Hint: Energy is conserved.

Answer 1

mass of lead m = 0.2 kg;

height h = 40 m;

specific heat of lead C = 1.28 x 10^2 J/kg-c;

(a) initial potential energy:

Total potential energy = Gravitational potential energy + internal energy

Gravitational potential energy = mgh;

= 0.2 x 9.8 x 40;

= 78.4 J.

Since, lead is at 15 degree celcius temperature, therefore its internal enegy;

dU = mCdT;

= 0.2 x 128 x 15;

= 384 J.

initial potential energy = 78.4 + 384;

= 462.4 J.

(b) Internal energy of the lead (ie 384 J) will lost as heat and the rest gravitational potential energy will convert into kinetic energy.