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3 years ago

Write Python expressions using s1, s2, and s3 and operators and * that evaluate to: (a) 'ant bat cod'

Engineering

1 answer:

jenyasd209 [6]3 years ago

5 0

Given that the correct question is

Write Python expressions using s1, s2, and s3 and operators + and * that evaluate to: (a) 'ant bat cod'

Answer:

s1 = 'ant'

s2 ='bat'

s3 ='cod'

#using string concatenation

s4 = s1 + ' ' + s2 + ' ' +s3

print(s4)

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A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the

Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})$

Using the Boyle equation we have

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})$

Where,

$C_p$ = Specific heat at constant pressure

$T_1$ = Initial temperature of gas

$T_2$ = Final temprature of gas

R = Universal gas constant

$v_1$ = Initial specific Volume of gas

$v_2$ = Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

$T_1 = T_2, v_2=v_1$

The equation would turn out as

$\Delta S = C_p ln1-ln1$

<em>Therefore the entropy change of the ideal gas is 0</em>

Into the surroundings we have that

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

$\Delta S = \frac{230kJ}{(30+273)K}$

$\Delta S = 0.76 kJ/K$

<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>

8 0

3 years ago

Decreasing unnecessary and wasteful purchases, decreasing the volume of waste, and decreasing

Feliz [49]

The answer is A
Hope this helps 8)

6 0

3 years ago

Suppose you must remove an average of 3.9×108J of thermal energy per day to keep your house cool during the summer. Part A If yo

noname [10]

Answer:

The required mechanical work is required to reduce each day by 1.05×10^8 Joules.

Explanation:

Coefficient of Performance (COP) = Q/W

Q is thermal energy absorbed by the air conditioner

W is mechanical work done

Q = 3.9×10^8 J

COP of old air conditioner = 2.3

W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J

COP of new air conditioner = 6

W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J

Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J

3 0

3 years ago

What is 220 C in degrees Fahrenheit (F)?

8_murik_8 [283]

Answer:

428°F

Explanation:

The equation to convert degrees Celsius to degrees fahrenheit is

°F (degrees fahrenheit) = (9/5 * °C (degrees celsius) ) + 32

°F = (9/5 * 220 °C (degrees celsius)) +32 = 428 °F (degrees fahrenheit)

4 0

3 years ago

Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What

Tresset [83]

Answer:

the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

Explanation:

Given that:

Initial Temperature $T_1$ = 15°C

Initial Pressure $P_1$ = 5 atm

Final Pressure $P_2$ = 1 atm

Data obtain from steam tables of saturated water at 15°C are as follows:

Specific volume v = 1.001 cm³/gm

The change in temperature = 2°C

Specific heat of water = 4.19 J/gm.K

volume expansivity β = 1.5 × 10⁻⁴ K⁻¹

The expression to determine the change in temperature can be given as :

$\delta \ T = \frac{-V (1- \beta \ T}{C_p} * \delta \ P ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$ $\delta \ T = \frac{-1.001 \frac{cm^3}{gm} (1- 1.5*10^{-4} \ K^{-1} )*2}{4.19 \ \frac{J}{gm.K}} *(5-1)atm ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$

Δ T = 0.093 K

Now; we can calculate the lost work bt the formula:

$W_{lost} = T_{surr} *S$

where ;

$T_{surr}$ is the temperature of the surrounding. = 20°C = (20+273.15)K = 293.15 K

From above the change in entropy is:

$\delta \ S = C_p \ In (\frac{T+ \delta \ T }{T}) * \beta V \delta P$

$\delta \ S = 4.19* \ In (\frac{288.15+0.093 }{288.15}) - 1.5*10^{-4} * 1.001 (5-1)* (\frac{1}{9.87})$

$\delta \ S =1.408*10^{-3} \ J/gm.K$

$W_{lost} = T_{surr} *S$

$W_{lost} = 293.15* 1.408*10^{-3} \ J/gm.K$

$W_{lost} = 0.413 \ kJ/kg$

Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

6 0

4 years ago