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Cerrena [4.2K]
2 years ago
7

Approximately 220 million tires are discarded in the U.S. each year. These tires present a disposal problem because they take up

space, harbor pests, and have been known to catch fire. One tire can generate about 250,000 BTUs (1 BTU = 3 x 10 -4 kWh) when it is burned. The average American home consumes about 10,000 kWh of electricity per year. How many tires would be needed to meet the annual electricity demand of ten homes for one year if the production of electricity from tires is 50% efficient?
Chemistry
1 answer:
Andreas93 [3]2 years ago
8 0

Answer:

2667 tires are needed to meet the demand of ten homes for one year.

Explanation:

According to the Second Law of Thermodynamics, only a part of generated energy when tires are burned can be utilized due to irreversibilities associated with finite temperature differences. The energy from a tire that can be transformed into electricity (E_{out}), measured in kilowatt-hours, is estimated by definition of efficiency:

E_{out} = \eta \cdot E_{in}

Where:

\eta - Efficiency, dimensionless.

E_{in} - Energy liberated by burning, measured in kilowatt-hours.

Given that \eta = 0.5 and E_{in} = 75\,kWh, the amount of energy per year generated by a tire is:

E_{out} = 0.5\cdot (75\,kWh)

E_{out} = 37.5\,kWh

Now, the amount of tires needed to meet the demand of then homes for one year is:

n = \frac{(10\,homes)\cdot \left(10000\,\frac{kWh}{home} \right)}{37.5\,\frac{kWh}{tire} }

n = 2666.667\,tires

2667 tires are needed to meet the demand of ten homes for one year.

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How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C
NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0
2 years ago
Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t
Dafna11 [192]

Answer:

The plane with aluminium can lift more mass of passangers than the plane of steel.

Explanation:

The total mass the airplane canc lift is:

m_{tot}=m_{fuselage}+m_{passangers}

For aluminium:

m_{tot}=m_{fus-Al}+m_{pas-Al}

m_{fus-Al}=\delta _{Al}*V_{fuselage}

and

V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]

where:

  • L is lenght
  • D is diameter
  • e is thickness

m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}

For steel (same procedure):

m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel

Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.

Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.

5 0
2 years ago
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
2 years ago
A newborn blue whale measures 6.0 to 7.9 m long and weighs up to 3003 kg. The blue whale produces the loudest sound by any livin
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3 0
2 years ago
Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0
REY [17]

Answer:

pH = 7.1581

Explanation:

The equilibrium of NaHSO₃ with Na₂SO₃ is:

HSO₃⁻ ⇄ SO₃²⁻ + H⁺

<em>Where K of equilibrium is the Ka2: 6.5x10⁻⁸</em>

HSO₃⁺ reacts with NaOH, thus:

HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺

As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:

HSO₃⁻: 0.252 moles

SO₃²⁻: 0.139 moles

Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:

0.0500L ₓ (1mol /L): 0.050 moles of NaOH.

Thus, final moles of both compounds are:

HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles

SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles

Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:

pH = pka + log [SO₃²⁻] / [HSO₃⁻]

Where pKa is - log Ka = 7.187

Replacing:

pH = 7.187 + log [0.189] / [0.202]

<h3>pH = 7.1581</h3>

4 0
2 years ago
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