A small lead ball with a volume of 0.01 cm3 falls evenly in the water. How much heat will be released

A small lead ball with a volume of 0.01 cm3 falls evenly in the water. How much heat will be released when the ball moves 6 m?

Given:

V = 0.01 cm3 = 0.01 * 10-6 m3 – the volume of the lead ball;

ro = 11300 kg / m3 is the density of lead;

g = 10 m / s2 – free fall acceleration;

L = 6 meters;

ro1 = 1000 kg / m3 – water density.

It is required to determine how much heat Q (Joule) will be released when the ball moves a distance L.

According to the condition of the problem, the ball falls uniformly in the water. Then, according to Newton’s first law:

F gravity + F resistance = Farchimedes;

Fresistance = F gravity – Farchimedes;

Fresistance = m * g – V * g * ro1 = V * ro * g – V * g * ro1 =

= V * g * (ro – ro1) = 0.01 * 10-6 * 10 * (11300 – 1000) = 0.01 * 10-5 * 10300 =

= 0.01 * 10-1 * 1.03 = 0.001 Newton.

The work of the resistance force will be spent on heating the ball. Then:

Q = A = Fresistance * L = 0.001 * 6 = 0.006 Joules = 6 mJ.

Answer: when the ball moves, a heat equal to 6 mJ will be released.