Final answer:
The temperature of a 5.00 kg lead sphere that is dropped from an 80.0-m-tall building will rise by 6.125 °C when its potential energy is converted to heat upon hitting the sidewalk.
Step-by-step explanation:
To calculate how much the temperature of a 5.00 kg lead sphere will rise when it is dropped from the top of an 80.0-m-tall building and all of its kinetic energy is converted into heat upon hitting the sidewalk, we use the principle of conservation of energy and the specific heat capacity of lead.
Calculations
The potential energy (PE) of the lead sphere at the top of the building is given by PE = mgh, where m is the mass of the sphere, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the building. This potential energy is converted into kinetic energy (KE) as the sphere falls and then into heat (Q) when it hits the sidewalk. The specific heat capacity (c) of lead is needed to determine the temperature change (ΔT).
First, we calculate the potential energy:
PE = mgh = (5.00 kg) × (9.8 m/s²) × (80.0 m) = 3920 J
Since all the kinetic energy is converted into heat:
Q = PE = 3920 J
Next, we use the formula Q = mcΔT to find ΔT, where m is the mass of the lead sphere, c is the specific heat capacity of lead (approximately 128 J/kg·°C), and ΔT is the change in temperature.
Rearranging for ΔT gives:
ΔT = Q / (mc) = 3920 J / (5.00 kg × 128 J/kg·°C)
ΔT = 6.125 °C
Therefore, the temperature of the lead sphere will rise by 6.125 °C when it hits the sidewalk.