In a lunar experiment, a 950-g aluminum (920 J/(°Ckg)) sphere is dropped from the space probe while is 75 m above the Lunar ground. If the sphere’s temperature increased by 0.11°C when it hits the ground, what percentage of the initial mechanical energy was absorbed as thermal energy by the aluminum sphere?

13.759 % of the initial mechanical energy is lost as thermal energy.

Explanation:

By the First Law of Thermodynamics we know that increase in internal energy of the object (), in joules, is equal to the lost amount of the change in gravitational potential energy (), in joules:

(1)

Where is the percentage of the energy loss, no unit.

By definition of the gravitational potential energy and internal energy, we expand this equation:

(1b)

Where:

- Mass of the object, in kilograms.

- Gravitational acceleration, in meters per square second.

- Initial height of the object above the lunar ground, in meters.

- Specific heat of aluminium, in joules per degree Celsius-kilogram.

- Temperature increase due to collision, in degree Celsius.

If we know that , , , and , then the percentage of energy loss due to collision is:

13.759 % of the initial mechanical energy is lost as thermal energy.