How many degrees will a piece of lead weighing 2 kg heat up when it falls to the ground
How many degrees will a piece of lead weighing 2 kg heat up when it falls to the ground from a height of 21.3 m, if all the mechanical energy of the lead turns into its internal energy?
m = 2 kg.
h = 21.3 m.
g = 10 m / s2.
C = 140 J / kg * ° C.
E = Q.
Δt -?
At the beginning of the fall, the total mechanical energy of the body consists only of the potential energy E = En. Let us express the value of potential energy by the formula: En = m * g * h, where m is the mass of the body, g is the acceleration of gravity.
The amount of heat energy that goes into heating the lead ball is expressed by the formula: Q = C * m * Δt, where C is the specific heat of lead, m is the mass of the ball, Δt is the change in its temperature.
Since by the condition of the problem E = Q, then m * g * h = C * m * Δt.
Δt = m * g * h / C * m = g * h / C.
Δt = 10 m / s2 * 21.3 m / 140 J / kg * ° C = 1.5 ° C.
Answer: The temperature of the ball will increase by Δt = 1.5 ° C.