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A 50-g block of copper (one mole has a mass of 63.5 g) at a temperature of 35C35^{\circ} \mathrm{C} is put in contact with a 100-g block of aluminum (molar mass 27 g) at a temperature of 20C.20^{\circ} \mathrm{C}. The blocks are inside an insulated enclosure, with little contact with the walls. At these temperatures, the high-temperature limit is valid for the specific heat. Calculate the final temperature of the two blocks. Do NOT look up the specific heats of aluminum and copper; you should be able to figure them out on your own.

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Answered 2 years ago
Answered 2 years ago
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If one quanta is equal to: q=3.61021021q = 3.6102 \cdot 10^{-21} J, then energy of 20 quanta is: E20=203.61021021=7.22041020E_{20} = 20 \cdot 3.6102 \cdot 10^{-21} = 7.2204 \cdot 10^{-20} J.

Entropy of this state is then:

S20=kBlnΩ=1.38071023ln(4.911026)S20=8.48561022J/K.\begin{align*} S_{20} &= k_B \ln \Omega \\ &= 1.3807 \cdot 10^{-23} \cdot \ln \left(4.91 \cdot 10^{26} \right) \\ S_{20} &= 8.4856 \cdot 10^{-22} \, \text{J/K.} \end{align*}

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