The figure below shows a \"self-cooling\" beverage can The can is equipped with

Question

The figure below shows a "self-cooling" beverage can The can is equipped with an outer jacket containing sodium carbonate (Na_2CO_3), which dissolves in water rapidly and endothermically. Na_2CO_3(s) rightarrow 2 Na^+(aq) + CO_3^2-(aq) Delta H degree = 67.7 kJ The user add, water to the outer jacket, and the heat absorbed in the chemical reaction chills the drink. The can contains 200 g of drink, the jacket contains 55 g of Na_2 CO_3(s) rightarrow 2 Na^+(aq) + CO_3^2-(aq) Delta H degree = 67.7 kJ The user adds water to the outer jacket, and the heat absorbed in the chemical reaction chills the drink. The can contains 200 g of drink, the jacket contains 55 g of Na_2 CO_3, and 100 g of water are to be added. If the initial temperature of the can and the water are 34 degree C on a summer day, what is the coldest temperature that the drink can reach? The can itself has a heat capacity of 40 J/degree C. Assume that the Na_2CO_3 solution and the drink both have the same heat capacity as pure water, 4.184 J g^-1 degree C^-1.

Explanation / Answer

We do an energy balance of the system:

Qgained = -Qlost

Qlost = mwaterCpdeltaT + msolutionCpdeltaT + CpcaldeltaT

Qlost = [200g * 4.184 J/gC * (Tf - 34)C] + [40 J/C * (Tf - 34)C] + [155g * 4.184 J/gC * (Tf - 34)]

Qgained = 67.7 kJ per mole

Number of moles of reactant: 55 g of Na2CO3 * (1mol / 106g) = 0.5189 moles

Qgained = 67.7 kJ / mol * 0.5189 moles = 35.13 kJ = 35130 J

Doing the balance:

35130 J = -([200g * 4.184 J/gC * (Tf - 34)C] + [40 J/C * (Tf - 34)C] + [155g * 4.184 J/gC * (Tf - 34)])

35130 J = -([836.8Tf - 28451.2] + [40Tf - 1360] + [648.52Tf - 22049]

35130 J = -([1525.32Tf - 51860.2])

35130 = -1525.32Tf + 51860.2

-16730.2 = -1525.32Tf

Tf = 10.9683 ºC

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.