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3 years ago

The speed of light is around 6.706×10^8 miles per hour. What is the speed of light in units of miles per minute?

Chemistry

1 answer:

Sindrei [870]3 years ago

7 0

<u>Answer:</u> The speed of light in miles per minutes is $1.117\times 10^7miles/min$

<u>Explanation:</u>

We are given the speed of light is $6.706\times 10^8miles/hr$ and we need to convert it into miles/min. So, we use the converion factor:

1 hour = 60 minutes

Converting that quantity into miles/minutes, we get:

$(\frac{6.706miles}{1hr})(\frac{1hr}{60min})=\frac{1.117\times 10^7miles}{min}$

Hence, the speed of light in miles per minutes is $1.117\times 10^7miles/min$

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Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$

The pressure is constant, so: $w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}$ (There is a multiplication by 100 due to the conversion of bar to kPa)

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$P_{1}V_{1}=nRT_{1}$

$P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}$

Subtracting the first from the second:

$1.5P_{1}V_{1}=nR(T_{2}-T_{1})$

Isolating $T_{2}$ :

$T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}$

Assuming that it is water steam, n=0.1666 kmol

$V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}$

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