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Lelechka [254]
3 years ago
12

A steam engine (assume a Carnot engine ) has an efficiency of 50.8%. If the waste heat has a temperature of 72.4 ◦C, what is the

temperature of the boiler? Answer in units of ◦C.
Chemistry
1 answer:
Inessa05 [86]3 years ago
3 0

Answer:

The temperature of the boiler is approximately 147.1 °C

Explanation:

A Carnot engine is an ideal engine that has the highest efficiency among all the engines because the second law of thermodynamics.That efficiency \eta  is:

\eta=1-\frac{T_{c}}{T_{h}}

with T_{h} the temperature of the hot reservoir (the boiler temperature) and T_{c} the temperature of the cold reservoir (the steam temperature). Solving for T_{h}:

T_{h}=\frac{-T_{c}}{\eta-1}=\frac{-72.4}{0.508-1}\approx147.1\,C

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A cylindrical specimen of some metal alloy having an elastic modulus of 126 GPa and an original cross-sectional diameter of 3.4
Delvig [45]

Answer:

The maximum length of the specimen before deformation is 240.64 mm

Explanation:

Strain = stress ÷ elastic modulus

stress = load ÷ area

load = 2130 N

diameter = 3.4 mm = 3.4×10^-3 m

area = πd^2/4 = 3.142 × (3.4×10^-3)^2/4 = 9.08038×10^-6 m^2

stress = 2130 N ÷ 9.08038×10^-6 m^2 = 2.35×10^8 N/m^2

elastic modulus = 126 GPa = 126×10^9 Pa

Strain = 2.35×10^8 ÷ 126×10^9 = 0.00187

Length = extension ÷ strain = 0.45 mm ÷ 0.00187 = 240.64 mm

5 0
3 years ago
2HgO—>2Hg+O2. I start with 46.8 g HgO. What is my theoretical yield of O2?
jok3333 [9.3K]

Answer:

Theoretical yield = 3.52 g

Percent yield =65.34%

Explanation:

Given data:

Mass of HgO = 46.8 g

Theoretical yield of O₂ = ?

Percent yield of O₂ = ?

Actual yield of O₂ =  2.30 g

Solution:

Chemical equation:

2HgO  →   2Hg + O₂

Number of moles of HgO = mass/ molar mass

Number of moles of HgO = 46.8 g / 216.6 g/mol

Number of moles of HgO = 0.22 mol

Now we will compare the moles of HgO with oxygen.

                         HgO             :             O₂

                             2               :               1

                            0.22           :          1/2×0.22 = 0.11 mol

Theoretical yield:

Mass of oxygen = number of moles × molar mass

Mass of oxygen =  0.11 mol × 32 g/mol

Mass of oxygen = 3.52 g

Percent yield :

Percent yield = actual yield / theoretical yield × 100

Percent yield = 2.30 g/ 3.52 g × 100

Percent yield =65.34%

5 0
3 years ago
4.14 Oxygen requirement for growth on glycerol Klebsiella aerogenes is produced from glycerol in aerobic culture with ammonia as
Iteru [2.4K]

Answer:

0.37 g

Explanation:

The molecular weight for Glycerol = 92

Number of Carbon atoms in glycerol (x)  C_3H_8O_3 = 3

Molecular weight of  the Biomass  ( Klebsiella aerogenes )

= CH_{1.73}O_{0.43}N_{0.24}

= \frac{23.97}{0.92}

= 26.1

From the molecular weight of the Biomass, we can deduce the Degree of reduction for the substrate(glycerol denoted as \delta _g) as follows:

= (4×1)+(1×1.73)-(2×0.43)-(3×0.24)

= 4.15

Given that the yield of the Biomass = 0.40 g

However;

C = Yield of Biomass *\frac{Molecular weight of substrate}{Molecular weight of the Biomass}

C = 0.40*\frac{92}{26.1}

C = 1.41 g

Now , the oxygen requirement can be calculated as:

= \frac{1}{4}*(n*S -  C * \delta _{g})

= \frac{1}{4}(3*4.7-1.41*4.15)

= 2.1 g/mol

Hence, we can say that the needed oxygen = 2.1 g/mol of the substrate consumed.

Now converting it to mass terms; we have:

= 2.1*\frac{number of mole of oxygen}{molecular weight of glycerol}

= 2.1 * \frac{16}{92}

= 0.3652 g

≅ 0.37 g

∴ The oxygen requirement for this culture in mass terms = 0.37 g

3 0
3 years ago
list examples of foliated and nonfoliated rocks explain the difference between the two metamorphic rocks
weqwewe [10]

Explanation:

Foliated metamorphic rocks such as gneiss, phyllite, schist and slate have a layered or banded appearance that is produced by exposure to heat. Non Foliated rocks such as hornfels, marble and quartzite does not have a layered or banded appearance

5 0
4 years ago
How does a catalyst appear in a chemical equation
Alinara [238K]
<span> A </span>catalyst<span> will </span>appear<span> in the steps of a </span>reaction<span> mechanism, but it will not </span>appear<span> in the overall </span><span>chemical reaction</span>
4 0
4 years ago
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