Acube of aluminum 21.0 cm on each side is cooled from 105 °c to 30 °c. if the energy removed from the aluminum cube were added to a copper cube of the same size at 40 °c, what would be the final temperature of the copper cube? (rhoal = 2700 kg/m³; rhocopper = 8890 kg/m³)

95.4ºC

Explanation:

It can be showed that the heat needed to remove from a body in order to lower his temperature, is proportional to the mass of the body, and to the temperature difference to achieve.

The proportionality constant is called specific heat capacity (c), and is different for each material.

So, we can write:

QAl =c. m. (T2 –T1)  

As we are talking about a cube, we need to find out the mass, relating the density and the volumen, as follows:

mAl = ρAl . V = ρAl l3 = 2,700 kg/m3 . (0.21)3 m3 = 25 kg

Once we have the mass, we can get the energy removed from the cube, taking into account that the specific heat for Al is 921.1 J/kg.ºC, as follows:

QAl = 921.1 J/kg.ºC . 25 kg. (30ºC-105ºC) = 1,727.1 KJ

This same energy, will be provided to an identical copper cube, so we need to replace the new values for the mass and the specific heat capacity:

mCu = ρCu . V = 8,890 kg/m3 . (0.21)3 m3 = 82.3 kg.

The specific heat capacity for copper, in J/kg.ºC, is equal to 376.8, so we can include these values in the formula for Q as follows:

QCu = 376.8 J/kg.ºC. 82.3 kg. (t2-40ºC)

We know that QCu, is equal in magnitude to QAl, so we can solve for T2:

T2 = 40ºC + (1,727 KJ / 376.8 J/kg.ºC .82.3 kg) = 95.4ºC

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