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anastassius [24]
3 years ago
7

A combination of a heat engine driving a heat pump (see Fig. P7.106) takes waste energy at 50°C as a source Qw1 to the heat engi

ne rejecting heat at 30°C. The remainder Qw2 goes into the heat pump that delivers a QH at 150°C. If the total waste energy is 5 MW find the rate of energy delivered at the high temperature.
Chemistry
1 answer:
erma4kov [3.2K]3 years ago
7 0

Answer:

1.08 MW

Explanation:

Given that:

Heat supplied to heat engine

Q_{WS} = 5 MW

Temperature of the hot source of engine

T_{WS} = 323 K

Temperature of the cold source of engine

T_L = 303 K

the temperature and heat relation for a reversible heat engine can be presented as follows:

\frac{T_L}{T_{WS}} =\frac{Q_L}{Q_{WS}}

\frac{303}{323} =\frac{Q_L}{5}

Q_L = 4.69 MW

TO determine the work done by the heat engine ; we have:

W_{engine}=Q_{WS}-Q_{L}

W_{engine}=5 -4.69

W_{engine}=0.309MW

The temperature and heat relation for the reversible heat pump can be written as:

\frac{T_H}{T_L}=\frac{Q_H}{Q_L}

\frac{T_H}{T_L}=\frac{Q_H}{Q_H-W}

\frac{423}{303} =\frac{Q_H}{Q_H-W}

Q_H=1.396-0.309\\Q_H=1.08MW

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A sample of an unknown gas at STP has a density of 0.630 gram per liter. What is the gram molecular mass of this gas?
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PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
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3 0
3 years ago
Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

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Dima020 [189]

Answer:

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