A .095kg aluminum sphere is dropped rom the roof of a 45m building.If 65% of the thermal energy produced when the sphere hits the ground is absorbed by the sphere, what is its temperature change?
To determine the temperature change of the aluminum sphere when it hits the ground, we need to calculate the thermal energy produced and then find 65% of that energy.
First, we need to determine the potential energy of the sphere when it is dropped from the roof. The potential energy (PE) is given by the equation PE = mgh, where m is the mass of the sphere (0.095 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height of the building (45 m).
PE = (0.095 kg) * (9.8 m/s²) * (45 m)
PE = 39.915 J
Next, we need to calculate the thermal energy produced when the sphere hits the ground. Assuming all the potential energy is converted to thermal energy (ignoring other energy losses like air resistance), the thermal energy (TE) can be calculated using TE = PE.
TE = 39.915 J
Now, we need to find 65% of the thermal energy absorbed by the sphere. We can calculate this by multiplying the thermal energy by 0.65.
Thermal energy absorbed = TE * 0.65
Thermal energy absorbed = 39.915 J * 0.65
Thermal energy absorbed = 25.94175 J
The temperature change of the sphere can be determined using the specific heat capacity (c) of aluminum. The formula for temperature change (ΔT) is given by the equation ΔT = (thermal energy absorbed) / (mass * specific heat capacity).
The specific heat capacity of aluminum is approximately 0.897 J/g°C, or 0.897 J/kg°C.
ΔT = (Thermal energy absorbed) / (mass * specific heat capacity)
ΔT = 25.94175 J / (0.095 kg * 0.897 J/kg°C)
ΔT ≈ 291.75°C
Therefore, the temperature change of the aluminum sphere when it hits the ground is approximately 291.75°C.
Ke = m g h = .095*9.8 * 45 Joules
.65 Ke = .045 (specific heat of Al)(delta T)