Final answer:
The temperature of the lead shot increases by 3.82°C after it is dropped from a height of one meter fifty times, based on the conservation of energy where the gravitational potential energy is entirely converted into thermal energy.
Step-by-step explanation:
To calculate the increase in temperature of the lead shot when dropped through a vertical distance of one meter fifty times, we use the principle of conservation of energy. The gravitational potential energy (GPE) lost is converted into thermal energy. The GPE lost for one drop is mgh, where m is the mass of the lead shot, g is the acceleration of gravity (9.8 N/kg), and h is the height (1 meter).
For 0.2 kg of lead dropped through 1 meter, the energy conversion for one drop is: GPE = mgh = 0.2 kg × 9.8 N/kg × 1 m = 1.96 J. When dropped fifty times, the total energy transferred to the lead is 50 × 1.96 J = 98 J. The increase in temperature (ΔT) can be found using the formula ΔT = Q / (mc), where Q is the total energy transferred, m is the mass, and c is the specific heat capacity of lead.
The increase in temperature is: ΔT = 98 J / (0.2 kg × 128 J/kg°C) = 3.82°C. Hence, the temperature of the lead increases by 3.82°C after being dropped fifty times.