1) Calculate the calorimeter constant, Ccal. Assume that 1.0 mL of water has a m

Question

1) Calculate the calorimeter constant, Ccal. 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. We placed 2 styrofoam cups together as a calorimeter and mixed 30 mL of hot water heated to 60 degrees celcius with 30 mL of water that was already in the calorimeter at room temp. (28°C) and the temp. of the hot water was 75°C. FInal temp. after the mixture was 44°C.

2) Calculate the value of sm (specific heat of a metal). 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.

3) Same experiment but now with acid and base mixture, Using the average Ccal from the previous week’s experiment 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 / 0C-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. Using Hess’ Law, and the data in Appendix 4 of your text, calculate the expected molar H of neutralization for each acid. 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 only 3 equations for the questions.... ]

qhw = qcw + qcal

sw x mhw x (Th – Tf) = sw x mcw x (Tf – Tc) + Ccal (Tf – Tc)

[note: qhw must be greater than qcw to account for heat lost to the calorimeter] then:

Ccal = (qhw – qcw) / (Tf – Tc) in units of (J / °C)

Explanation / Answer

1) Calculate the calorimeter constant, Ccal.

- 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.

We placed 2 styrofoam cups together as a calorimeter and mixed 30 mL of hot water heated to 60 degrees celcius with 30 mL of water that was already in the calorimeter at room temp. (28°C) and the temp. of the hot water was 75°C. FInal temp. after the mixture was 44°C.

Qcal + Qhot + Qcool = 0

Qcal = -(Qhot + Qcool)

Qcal = Ccal*(Tf-th)

Qhot = mhot*Cp*(Tf-Th)

Qcool = mcool *Cp*(Tf-Tc)

substitute data

Qcal = -(Qhot + Qcool)

Ccal*(Tf-Tcal) = -(mhot*Cp*(Tf-Th) + mcool *Cp*(Tf-Tc))

Ccal*(44-60) = -(30*4.184*(44-75) + 30*4.184*(44-28))

solve

Ccal = (1882.8 ) /((44-60)

Ccal = 117.675 J/°C

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