1 part A) Calculate the value of sm (specific heat of a metal) knowing that the

Question

1 part A) Calculate the value of sm (specific heat of a metal) knowing that the Ccal [calorimeter constant is 117.675 J/°C]. Assume that 1.0 mL of water has a mass of 1.0 g. The specific heat for water is 4.184 J / C-g. The mass of the metal was 83.274 grams, volume of cold water (28°C) was 35 mL, temp of hot water and hot metal when placed in hot water was 80°C, the metal was then quickly moved to cold water in the calorimeter and final temp was 35°C. The density of the metal was 9.270 g/mol.

1 Part B) If we were to do the same experiment but now with acid and base mixture, using the Ccal you have found, and calculate the H for each acid neutralization reaction above. Assume that 1.0 mL of acid or base solution used here has a mass of 1.0 g. The specific heat for water is 4.184 J / C-g. From the H calculated above and the number of moles of acid neutralized in each case, calculate the molar heat of neutralization for HCl and H2SO4. In each case also write the balanced chemical equation for neutralization of each by aqueous NaOH.

We poured 40 mL 1.0 M of HCl with 20 mL of 2.0 M NaOH in the calorimeter as the temp of the acid was 23.4°C and the base temp was 23.5°C. Final temp in the end was 31.9°C. Then we replaced HCl with 20 mL H2SO4 and mixed it with the base again fresh.

[We were given 3 equations ]

qhw = qcw + qcal
sw x mhw x (Th – Tf) = sw x mcw x (Tf – Tc) + Ccal (Tf – Tc)
then: Ccal = (qhw – qcw) / (Tf – Tc) in units of (J / °C)

Explanation / Answer

Part A)

Heat gain by water (qwater) = mCpdT

                                             = 35 x 4.184 (35 - 28) = 1025.08 J

Specific heat of metal = 1025.08/83.274 (80 - 35)

                                   = 0.273 J/C.g

Part B) dH (HCl + NaOH) = mCpdT/(0.020 l x 2 M) mol

                                         = 60 x 4.184 (31.9 - 23.45)/0.04 x 1000

                                         = 53.032 kJ/mol

dH (H2SO4 + NaOH) = 40 x 4.184 (31.9 - 23.45)/1000 x 0.04 mol

                                   = 35.355 kJ/mol

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