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worty [1.4K]
3 years ago
7

Find the initial temperature of Aluminum if 35 g of it produces 781.2 Joules at the final temperature of 44.3 ºC.

Chemistry
1 answer:
Annette [7]3 years ago
6 0

Answer:

T initial =  19.55°C

Explanation:

Given data:

Mass of Al = 35 g

Energy produces = 781.2 J

Final temperature = 44.3°C

Initial temperature = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of Al is 0.902 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

Q = m.c. ΔT

781.2 J = 35 g×0.902 J/g.°C× (T final - T initial)

781.2 J = 35 g×0.902 J/g.°C× (44.3°C -T initial )

781.2 J = 31.57 j/°C× (44.3°C -T initial )

781.2 J/ 31.57 j/°C = (44.3°C -T initial )

24.75°C = (44.3°C -T initial )

24.75° - 44.3°C = -T initial

T initial =  19.55°C

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Which of the following solutions is a good buffer system?A solution that is 0.10 M HCN and 0.10 M NaClA solution that is 0.10 M
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A solution that is 0.10 M HCN and 0.10 M LiCN

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  • In this case; A solution that is 0.10 M HCN and 0.10 M LiCN, would make a good buffer system.
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Explanation:

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              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

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                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

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                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

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