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FromTheMoon [43]
3 years ago
7

Bromine, a liquid at room temperature, has a boiling point

Physics
1 answer:
lukranit [14]3 years ago
3 0

Yes it does !  The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided.  The boiling point is higher than room temperature.

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A bicycle rim has a diameter of 0.65 m and a moment of inertia, measured about its center, of 0.21 kg⋅m2What is the mass of the
stepan [7]

Answer:

m = 1.99 kg = 2 kg

Explanation:

The moment of inertia of a bicycle rim about it's center is given by the following formula:

I = mr^{2}\\

where,

I = Moment of Inertia of the Bicycle Rim = 0.21 kg.m²

r = Radius of the Bicycle Rim = Diameter of the Bicycle Rim/2

r = 0.65 m/2 = 0.325 m

m = Mass of the Bicycle Rim = ?

Therefore,

0.21\ kg.m^{2} = m(0.325\ m)^{2}\\m = \frac{0.21\ kg.m^{2}}{(0.325\ m)^{2}}\\

<u>m = 1.99 kg = 2 kg</u>

3 0
2 years ago
The magnitude of the momentum of an object is 64 kilogram meters per second. If the magnitude of te velocity is doubled, the mag
dsp73

Answer:

128 \frac{kg*m}{s}

Explanation:

P=m*v\\if double velocity:\\P=m*2v=2m*v\\

7 0
3 years ago
A heat engine does 9200 J of work per cycle while absorbing 22.0 kcal of heat from a hightemperature reservoir. What is the effi
Morgarella [4.7K]

Answer: 9.9%

Explanation: efficiency = (work output /work input) × 100

Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.

Work output = 9200 j and work input = 92048 j

Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%

4 0
3 years ago
According to the text, there is no energy shortage now, nor will there ever be. what reason (s) is given to support this stateme
Bas_tet [7]
The reason why there is no energy shortage nor will there ever be is because energy is being preserved and conserved and only changes form. It never gets lost or increased.
8 0
2 years ago
What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

4 0
3 years ago
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