In a lunar experiment, a 950-g aluminum (920 J/(degCkg)) sphere is dropped from the space probe while is 75 m above the Lunar ground. If the sphere’s temperature increased by 0.11degC when it hits the ground, what percentage of the initial mechanical energy was absorbed as thermal energy by the aluminum sphere?
To find the percentage of the initial mechanical energy that was absorbed as thermal energy by the aluminum sphere, we need to calculate the initial mechanical energy and the change in thermal energy.
First, let's calculate the initial mechanical energy:
Initial Mechanical Energy = Potential Energy
Potential Energy = mgh (mass x acceleration due to gravity x height)
Mass of the sphere = 950 g = 0.950 kg
Acceleration due to gravity on the Moon = 1.6 m/s^2
Height of the sphere above the Lunar ground = 75 m
Plugging in these values, we can calculate the initial potential energy:
Initial Mechanical Energy = 0.950 kg x 1.6 m/s^2 x 75 m
Next, we need to calculate the change in thermal energy:
Change in Thermal Energy = mcΔT (mass x specific heat capacity x change in temperature)
Mass of the sphere = 950 g = 0.950 kg
Specific heat capacity of aluminum = 920 J/(degCkg)
Change in temperature = 0.11 degC
Plugging in these values, we can calculate the change in thermal energy:
Change in Thermal Energy = 0.950 kg x 920 J/(degCkg) x 0.11 degC
Finally, we can calculate the percentage of initial mechanical energy absorbed as thermal energy:
Percentage = (Change in Thermal Energy / Initial Mechanical Energy) x 100
Substituting the calculated values:
Percentage = (0.950 kg x 920 J/(degCkg) x 0.11 degC) / (0.950 kg x 1.6 m/s^2 x 75 m) x 100
Simplifying the expression, we can find the answer by substituting the values and performing the calculations.