The average Olympic swimming pool will hold 2.5 million liters of water. If I wa

Question

The average Olympic swimming pool will hold 2.5 million liters of water. If I want to raise the temperature of the pool by 4degree C, how much energy in the form of heat must I introduce into the water? For this problem, you can assume the density of the water to be 1.0 g/mL and the specific heat of the water to be 4.184 J/g-degree C. Show all of your work. If the power company charges me $0,085 for every 3.6 times 106 Joules of energy, how much does it cost for me to raise the temperature of the pool one degree? Ammonium nitrate will undergo an explosive decomposition in the presence of excess heat. Using the change in enthalpy associated with the following reaction, determine how much energy would be released during the decomposition of 1.0 kg of solid ammonium nitrate. Show all of your work. 2NH4NO3(S) 2N2(g) + O2(G) + 4H2O(l) deltarxnH = -412.2 kj

Explanation / Answer

6)

we know that

mass = density x volume

given

density = 1 g /ml

volume = 2.5 million liters = 2.5 x 10^6 L

volume = 2.5 x 10^9 ml

so

mass of water = 1 x 2.5 x 10^9

mass of water = 2.5 x 10^9 g

now

we know that

heat = mass x specific heat x temp change

so

heat = 2.5 x 10^9 x 4.184 x 4

heat = 4.184 x 10^10

so

4.184 x 10^10 joules of energy is required

b)

now

3.6 x 10^6 J ---> $ 0.085

4.184 x 10^10 J ----> $ y

y = 0.085 x 4.184 x 10^10 / 3.6 x 10^6

y = 987.88

so

the cost is $ 987.88

7)

we know that

moles = mass / molar mass

also

molar mass of NH4N03 = 80

so

moles of NH4N03 = 1000 / 80

moles of NH4N03 = 12.5

now

consider the reaction

2 NH4N03 --> 2N2 + 02 + 4H20    dH = -412.2 kJ

so

2 moles of Nh4N03 ---->   412.2 kJ

12.5 moles of Nh4N03 ----> y kJ

y = 412.2 x 12.5 / 2

y = 2576.25

so

2576.25 kJ of energy would be released for 1 kg of NH4N03 decomposition

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