How much the temperature of the lead ball that fell from 1 meter changed. After the impact, the ball remained motionless.

Given:

h = 1 meter – the height from which the lead ball fell;

g = 10 N / kg (Newton per kilogram) – acceleration due to gravity (rounded off);

c = 130 J / (kg * C) – specific heat of lead.

It is required to determine dt (degree Celsius) – how much the temperature of the ball will change when falling from a height of h.

Since the condition of the problem is not specified, we assume that the entire mechanical energy of the ball will transform into thermal energy upon impact. Then, according to the law of conservation of energy:

Epotential = Q;

m * g * h = c * m * dt, where m is the mass of the lead ball;

g * h = c * dt;

dt = g * h / c = 10 * 1/130 = 10/130 = 1/13 = 0.08 ° Celsius (the result has been rounded to one hundredth).

Answer: The temperature of the lead ball will increase by 0.08 ° Celsius.