A 0.095-kg aluminium sphere is dropped from the roof of a 55-m-high building. If 65% of the thermal energy produced when it hits the ground is absorbed by the sphere, what is its temperature increase?
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This is Giancoli Answers with Mr. Dychko. The energy produced by this fall of the aluminum sphere is equal to its final kinetic energy, and that final kinetic energy equals the initial potential energy that it has at the top of the building. So, the energy produced then is going to be mgh, initial potential energy mass of the sphere of 0.095 kilograms times 9.8 meters per second squared times the 55 meters of the fall which is 51.205 joules. The energy absorbed by the aluminum sphere is going to be 65% of that. So, 0.65 times 51.205 joules is 33.2833 joules. And to find the temperature change that'll happen. Well, we'll say the energy absorbed by the aluminum sphere is going to be its mass times the specific heat times the change in temperature. So, we divide both sides by mc. And the change temperature is going to be Q over mc. So, that's 33.2833 joules absorbed divided by 0.095 kilograms times 900 joules per kilogram Celsius degree, specific heat of aluminum. And that's going to be 0.39 Celsius degrees. So, it's going to have a 0.39 Celsius degree rise in temperature.
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