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3 years ago

Convert Gravitational constant (G) = 6.67×10^-11 Nm²kg^-2 to cm³ g ^-1 s^-2.

Physics

1 answer:

GaryK [48]3 years ago

3 0

Answer:

6.67×10⁻⁸ cm³/g/s²

Explanation:

6.67×10⁻¹¹ Nm²/kg²

= 6.67×10⁻¹¹ (kg m/s²) m²/kg²

= 6.67×10⁻¹¹ m³/kg/s²

= 6.67×10⁻¹¹ m³/kg/s² × (100 cm/m)³ × (1 kg / 1000 g)

= 6.67×10⁻⁸ cm³/g/s²

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An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s

gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

$Q = ms\Delta T + mL$

here we have

$Q = 40(950)(680 - 32) + 40(450 \times 10^3)$

$Q = 24624 kJ + 18000 kJ$

$Q = 42624 kJ$

Since this is the amount of aluminium per hour

so power required to melt is given by

$P = \frac{Q}{t}$

$P = \frac{42624}{3600} kW$

$P = 11.84 kW$

Since the efficiency is 85% so actual power required will be

$P = \frac{11.84}{0.85} = 13.93 kW$

Part B)

Total energy consumed by the furnace for 30 hours

$Energy = power \times time$

$Energy = 13.93 kW\times 30 h$

$Energy = 417.9 kWh$

now the total cost of energy consumption is given as

$R = P \times 20 \frac{Cents}{kWh}$

$R = 417.9 kWh\times 20 \frac{cents}{kWh}$

$R = 8357.6 Cents$

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3 years ago

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Answer:

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What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?

Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

heat of fusion of water = 334 J/g

heat of vaporization of water = 2257 J/g

specific heat of ice = 2.09 J/g·°C

specific heat of water = 4.18 J/g·°C

specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C

H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ

Heat energy required to change water from 100°C to steam at 100°C

H₄ = mL = 1000 x 2257 = 2257 kJ

Total heat energy required

H = H₁ + H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

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3 years ago

A gas is compressed at constant temperature from a volume of 5.68 L to a volume of 2.35 L by an external pressure of 732 torr. C

Vlada [557]

Answer: The work done in J is 324

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

$W=-P\Delta V=-P(V_2-V_1)$

W = amount of work done = ?

P = pressure = 732 torr = 0.96 atm (760torr =1atm)

$V_1$ = initial volume = 5.68 L

$V_2$ = final volume = 2.35 L

Putting values in above equation, we get:

$W=-0.96atm\times (2.35-5.68)L=3.20L.atm$

To convert this into joules, we use the conversion factor:

$1L.atm=101.33J$

So, $3.20L.atm=3.20\times 101.3=324J$

The positive sign indicates the work is done on the system

Hence, the work done for the given process is 324 J

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3 years ago

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IrinaK [193]

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