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This is the second of our many quizzes on Energy. Click on this to access more such competitive quizzes on General Chemistry.

Energy Quiz 2

chemistry

 What is the chemical formula of Potential Energy: 

 Potential energy is associated with objects which are present at some height from ground level.

 $\text{Potential energy of an object is defined as the product of mass of that object,} \\ \text{gravitational force and height at which that object is present,} \\ \text{i.e. P.E. }= mgh.$

 What is the correct chemical equation to calculate the Gibbs free energy?

 Gibbs free energy is a kind of thermodynamic energy which is calculated at constant temperature and pressure.

 $\text{Change in Gibbs free energy for any reaction is} \\ \text{calculated by subtracting the product of temperature} \\ \text{and change in entropy from change in enthalpy,} \\ \text{i.e. } \Delta G = \Delta H - T\Delta S$.

 What will be the height of a wall where an object of mass ${{45kg}}$ is placed whose potential energy is ${{1000J}}$?

 Potential energy is calculated by formula: P.E. = $mgh$.

 $\text{Given that, mass of object }= {{45kg}} \\ \text{Height at which object is placed = to find?} \\ \text{P.E. of object }= {{1000J}} \\ \text{As we know that P.E. is calculated as ‘mgh’,} \\ \text{and from this formula height will be calculated as follow:} \\{{h = }}\dfrac{{{{P}}{{.E}}{{.}}}}{{{{mg}}}} \\ \text{Here }{{g = 9}}{{.8m/}}{{{s}}^{{2}}} \text{ or } {{10m/}}{{{s}}^{{2}}} \\ {{h = }}\dfrac{{{{1000}}}}{{{{45 \times 10}}}}{{ = 2}}{{.2m}}$

 What will be the mass of a ball which is placed at ${{10m}}$ height from the ground, whose kinetic energy is ${{120J}}$ and total amount of energy involved in this displacement is ${{456J}}$?

 P.E. will be calculated from the K.E. & total energy and then from the P.E. mass will be calculated.

 $\text{Given that, K.E. of ball }= {{120J}} \\ \text{Total energy of ball }= {{456J}} \\ \text{Height at which ball is placed }= {{10m}} \\ \text{Total energy = P.E. + K.E.} \\ {{P}}{{.E}}{{. = 456 - 120 = 336J}} \\ \text{Now we calculate mass of the ball from the following formula:} \\ {{P}}{{.E}}{{. = mgh}} \\ \text{Or } {{m = }}\dfrac{{{{P}}{{.E}}{{.}}}}{{{{gh}}}} \\ {{m = }}\dfrac{{{{336}}}}{{{{10 \times 10}}}}{{3}}{{.36kg}}$

 Rahul raised a ball of mass ${{60g}}$ to a stairs set of height ${{20m}}$ above from the ground. Find the potential energy of the ball at the given height and find the kinetic energy of this ball at the half way down when Rahul pulls the ball towards the ground. Also calculate the velocity of the falling ball. <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b04673ff-c053-4b5a-b98c-1f57eb03c8c66397484364055272205.png"/>

 At the halfway potential energy is equal to the kinetic energy.

 $\text{Given that, mass of ball }= {{60g}} \\ \text{Height at which ball is placed }= {{20m}} \\ \text{Potential energy of ball at the height }= mgh \\ \text{Here; }{{g = 9}}{{.8m/}}{{{s}}^{{2}}} \text{ or } {{10m/}}{{{s}}^{{2}}} \\ {{P}}{{.E}}{{. = 60 \times 10 \times 20 = 12000J}} \\ \text{Potential energy of the ball at the half way when ball falls down} \\ = {{60 \times 10 \times 10 = 6000J}} \\ \text{And from the law of conservation of energy,} \\ \text{at the halfway P.E. = K.E.} \\ \text{So, K.E. of ball }= 6000J \\ \text{As we know that K.E } = \dfrac{{{1}}}{{{2}}}{{m}}{{{v}}^{{2}}} \\ \text{Velocity of the ball will be calculated from K.E. as follow:} \\ {{v = }}\sqrt {\dfrac{{{{2 \times K}}{{.E}}{{.}}}}{{{m}}}} \\  {{v = }}\sqrt {\dfrac{{{{2 \times 6000}}}}{{{{60}}}}} {{ = 14}}{{.14m/s}}$