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MakcuM [25]
3 years ago
12

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

r at a temperature of 23.0°C. After stirring, the final temperature of both copper and water is 26.0°C. Assuming no heat losses, and that the specific heat capacity of water is 4.184 J/(g·°C), what is the molar heat capacity of aluminum, Cm(Al)?
Physics
1 answer:
vivado [14]3 years ago
5 0

Answer: The molar heat capacity of aluminum is 25.3J/mol^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of water = 130.0 g

m_2 = mass of aluminiunm = 23.5 g

T_{final} = final temperature = 26.0^oC=(273+26)K=299K

T_1 = temperature of water = 23^oC=(273+23)K=296K

T_2 = temperature of aluminium = 100^oC=273+100=373K

c_1 = specific heat of water= 4.184J/g^0C

c_2 = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]

c_2=0.938J/g^0C

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity =0.938J/g^0C\times 27g/mol=25.3J/mol^0C

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An ideal gas with k 5 1.4 is flowing through a nozzle such that the Mach number is 1.8 where the flow area is 36 cm2. Approximat
kakasveta [241]

Answer:

The flow area at the location where the Mach number is 0.9 is 25.24 cm²

Explanation:

Here we have for isentropic flow;

\frac{A}{A^*} = \frac{1}{M}(\frac{2}{k+1} (1+\frac{k-1}{2}M^2))^{(\frac{k+1}{2(k-1)} )

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Therefore, plugging the values we get

\frac{36}{A^*} = \frac{1}{1.8}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}1.8^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.439

Therefore, A* = 36/1.439 = 25.01769 cm²

Where the Mach number is 0.9, we have

\frac{A}{25.02} = \frac{1}{0.9}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}0.9^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.009

Therefore A = 25.020× 1.009 = 25.24 cm²

The flow area at the location where the Mach number is 0.9 = 25.24 cm².

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