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3 years ago

Help ;-; Will give Brainliest and 25 Points

Chemistry

2 answers:

Alika [10]3 years ago

7 0

Hello there lol :))))

Yuri [45]3 years ago

3 0

I don’t see what you need help with but thanks:)

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Two Carnot engines are operated in series with the exhaust (heat output) of the first engine being the input of the second engin

OlgaM077 [116]

Answer:

(a) 140 F

(b) The temperature rise at the point where the heat is dumped is 2.51 degC

Explanation:

(a) Considering T1 the temperature of input of the first engine, T2 the temperature of the exhaust of the first engine (and input of the second engine) and T3 the exhaust of the second engine, if both engines have the same efficiency we have:

$\eta=1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}$

The temperatures have to be expressed in Rankine (or Kelvin) degrees

$1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}\\\\\frac{T_1}{T2}=\frac{T_2}{T_3}\\\\(T_2)^{2} =T_1*T_3\\\\T_2=\sqrt{T_1*T_3} =\sqrt{(459.67+260)*(459.67+40)}= \sqrt{719.67*499.67}\\\\ T_2=599 \, R= (599-459.67) ^{\circ} F=140^{\circ} F$

(b) The Carnot efficiency of the cycle is

$\eta_{c}=1-Th/Ts=1-(273+20)/(273+315)=0.502$

If the efficiency of the plant is 60% of the Carnot efficiency, we have

$\eta=0.6*\eta_{c}=0.6*0.502=0.302$

The heat used in the plant can be calculated as

$Q_i=W/\eta=750MW/0.302=2483MW$

And the heat removed to the heat sink is

$Q_o=Qi-W=2483-750=1733MW$

If the flow of the river is 165 m3/s, the heat per volume in the sink is

$\frac{Q_o}{f} =\frac{1733 MJ/s}{165 m3/s}= 10.5MJ/m3$

Considering a heat capacity of water C=4.1796 kJ/(kg*K) and a density ρ of 1000 kg/m3, the temperature rise of the water is

$\Delta Q=C*\Delta T\\\Delta T=(1/C)*\Delta Q\\\Delta T=(\frac{1}{4.1796\frac{kJ}{kgK} } )*10,500\frac{kJ}{m3}*\frac{1m3}{1000kg}\\\Delta T= 2.51 ^{\circ}C$

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4 years ago

How does the energy of an electron change when the electron moves closer to the nucleus?

tankabanditka [31]

It increases. B.

Close to the nucleus are high energy levels, because of the pull of the nucleus. But farther away from the nucleus, the electrons there are of low energy level.

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4 years ago

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DiKsa [7]

Answer:

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3 years ago

Which one will diffuse faster Ammonia or carbon dioxide Why?<br><br>

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jcp-fnoa-kzk join here

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3 years ago

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A 52 gram sample of an unknown metal requires 714 Joules of energy to heat it from

ratelena [41]

Answer: Approximately $0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}$

===================================================

Work Shown:

We have the following variables

Q = 714 joules = heat required
m = 52 grams = mass
c = specific heat = unknown
$\Delta t$ = 82-30.5 = 51.5 = change in temperature

note: the symbol $\Delta$ is the uppercase Greek letter delta. It represents the difference or change in a value.

Apply those values into the formula below. Solve for c.

$Q = m*c*\Delta t\\\\714 = 52*c*51.5\\\\714 = 52*51.5*c\\\\714 = 2678*c\\\\2678*c = 714\\\\c = \frac{714}{2678}\\\\c \approx 0.26661687826737\\\\c \approx 0.267\\\\$

The specific heat of the unknown metal is roughly $0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}$

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3 years ago