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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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No of moles:-

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\\ \tt\longmapsto \dfrac{7.62(10^{25})}{6.022(10^{23})}

\\ \tt\longmapsto 126.5mol

8 0
2 years ago
How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas
IrinaVladis [17]

94.6 g.  You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the <em>dilution formula</em>

<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2

where

<em>m</em> represents the mass and

<em>C</em> represents the percent concentrations

We can rearrange the formula to get

<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)

<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %

<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %

∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

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What is the identity of an ion having 46 protons, 42 electrons and 60 neutrons?
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Answer:

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