Question

The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.

What will be the equilibrium temperature when a 215 g block of copper at 245 ∘C is placed in a 155 g aluminum calorimeter cup containing 815 g of water at 16.0 ∘C?

Answer #1

Convert all the masses in kilograms -

Mass of Cu = 0.215 kg

Mass of Al = 0.155 kg

Mass of water, H2O = 0.815 kg

For Cu, Q = 0.215 * 390 * (245 – Tf) = 20543.3 – 83.8 * Tf

The initial temperature of aluminum calorimeter cup and water is 16
deg C

For Al, Q = 0.155 * 900 * (Tf – 16) = 139.5 * Tf – 2232

For water, Q = 0.815 * 4186 * (Tf – 16) = 3411.6 * Tf – 54585.4

So, total Q = 3551.1 * Tf – 56817.4

Therefore,

Loss in temperature by copper = Gain in temperature by aluminium calorimeter + water

=> 20543.3 – 83.8 * Tf = 3551.1 * Tf – 56817.4

=> 77360.7 = 3634.9 * Tf

=> Tf = 21.28 deg C

So, the equilibrium temperature of the mixture = 21.28 deg C. (Answer)

answered by: anonymous

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