Question

A thermally isolated system is made up of a hot piece of aluminum and a cold piece of copper; the aluminum and the copper are in thermal contact. The specific heat of aluminum is more than double that of copper. Which object experiences the greater temperature change during the time the system takes to reach thermal equilibrium?

Answer 1

Answer:

• Copper is the object that experiences the greater temperature change.

Explanation:

Thermally isolated system means that the system does not exchange thermal energy with the surroundings.

Hence, any thermal exchange, in virtue of the temperature difference of the aluminum and copper pieces, is between them.

In consequence, the law of conservation of energy states that the heat lost by the hot substance will be gained by the cold matter.

In equations, that is:

• Heat lost by aluminum = heat gained by copper.

Now, the gain or loss or heat of a substance, Q, is related with the mass (m), the specific heat (Cs), and the cahnge of temperature (ΔT), per the equation:

• Q = m × Cs × ΔT

∴ Q lost by aluminum = Q  gained copper ⇒

• [m × Cs × ΔT ] aluminum = [m × Cs × ΔT ] copper.

Under the reasoning assumption that the masses of aluminum and copper are equal, the equations is simplified to:

• [Cs × ΔT ] aluminum = [Cs × ΔT ] copper.

• Cs aluminum / Cs copper = ΔT copper / ΔT aluminum

• Cs aluminum > 2 × Cs copper  ⇒ Cs

• Cs aluminum / Cs copper > 2 ΔT copper / ΔT aluminum

• ΔT copper / ΔT aluminum > 2

• ΔT copper > 2 × ΔT aluminum

In words, since it is stated that  the specific heat of aluminum is more than double that of copper,  in order to keep the equality, ΔT of copper shall be more than double ΔT of aluminum.

Hence, the conclusion is that the object that experiences the greater temperature change is copper (the one with the lower specific heat), under the assumption that both objects have the same amount of matter (mass).