A 13.1-g sample of CaCl2 is dissolved in 104 g of water, with both substances at 24.7°C. Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 J °C-1 g-1.

Question

Chemistry

Everett Villanueva

9 May, 12:26

A 13.1-g sample of CaCl2 is dissolved in 104 g of water, with both substances at 24.7°C. Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 J °C-1 g-1.

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Susan Morse · Answer 1

The final temperature of the solution is 44.8 °C

Explanation:

assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:

Q dis + Q sol = 0

Using tables, can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C) is q dis = - 83.3 KJ/mol. And the molecular weight is

M = 1*40 g/mol + 2 * 35.45 g/mol = 110.9 g/mol

Q dis = q dis * n = q dis * m/M = - 83.3 KJ/mol * 13.1 g/110.9 gr/mol = - 9.84 KJ

Qdis = - 9.84 KJ

Also Qsol = ms * Cs * (T - Ti)

therefore

ms * Cs * (T - Ti) + Qdis = 0

T = Ti - Qdis * (ms * Cs) ^-1 = 24.7 °C - (-9.84 KJ/mol) / [ (104 g + 13.1 g) * 4.18 J/g°C] * 1000 J/KJ

T = 44.8 °C