Tag: the essence of change

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Questions Related to the essence of change

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A hammer of mass 1$\mathrm { kg }$ having speed of 50$\mathrm { m } / \mathrm { s }$ , hit a iron nail of mass 200$\mathrm { gm }$ . If specific heat of iron is 0.105 cal/gm'C and half the energy is converted into heat, the raise in temperature of nail is

  1. $7.1 ^ { \circ } C$
  2. $9.2 ^ { \circ } \mathrm { C }$
  3. $10.5 ^ { \circ } C$
  4. $12.1 ^ { \circ } \mathrm { C }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial kinetic energy = 0.5 * m * v^2 = 0.5 * 1 * 50^2 = 1250 J. Half of this is 625 J. Convert to calories: 625 / 4.2 = 148.8 cal. Heat = ms(delta T) -> 148.8 = 200 g * 0.105 cal/g C * delta T. delta T = 148.8 / 21 = 7.08 C.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

In case of explosion of a bomb which of the following does not change?

  1. Chemical energy

  2. Total energy

  3. Kinetic energy

  4. Mechanical energy
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The chemical energy possessed by the bomb after explosion converts into heat energy, sound energy, light energy i.e. into different forms of energy but remains conserved. Hence, total energy of bomb does not change, but kinetic and mechanical energy changes.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

The specific heat capacity of a metal at low temperautre (T) is given as $C _ { p } \left( k \sqrt { k } - 1 k g ^ { - 1 } \right) = 32 \left( \frac { T } { 400 } \right) ^ { 3 }$ A 100 gram vessel of this metal is to be cooled from $20 ^ { \circ } K$ to $4 ^ { \circ } \mathrm { K }$ by a special refrigerator operating at room temperature $\left( 27 ^ { \circ } \mathrm { C } \right) .$ The amount of work required to cool the vessel is:-

  1. equal to 0.002$\mathrm { kJ }$
  2. greater than 0.148$\mathrm { kJ }$
  3. between 0.148$\mathrm { kJ }$ and 0.028$\mathrm { kJ }$
  4. less than 0.028$\mathrm { kJ }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

An ideal gas heat engine operates in a Carnot's cycle between $227^{\circ}C$ and $127^{\circ}C$. It absorbs $6\times 10^{4} J$ at high temperature. The amount of heat converted into work is _____.

  1. $4.8\times 10^{4}J$
  2. $3.5\times 10^{4}J$
  3. $1.6\times 10^{4}J$
  4. $1.2\times 10^{4}J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Efficiency of Carnot engine = 1 - (T_low / T_high) = 1 - (400 / 500) = 0.2. Work done = Efficiency * Heat absorbed = 0.2 * 6 * 10^4 = 1.2 * 10^4 J.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

Work done by 100 calorie of heat is __________.

  1. 418.4 J

  2. 4.184 J

  3. 41.84 J

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
For an isothermal expansion of an ideal gas, the change in internal energy is zero.

According to the first law of thermodynamics, 

Change in internal energy U = Q-W = 0

So, all the heat energy is utilized to do work. 

Q = W

We know that, one calorie is equal to 4.184 J

Therefore, Work done by 100 calorie of heat in an isothermal expansion of any ideal gas will be 4.184 * 100 =  418.4 Joule


Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A person of weight 60 kg wants to loose 5 kg by going up and down 10m high stairs. Assume he burns twice as much fat while going up than going down. If 1 kg of fat is burnt on expending 7000 kcal. How many times must he go up and down to reduce his 7 kg weight? (Take $  g=10 \mathrm{ms}^{-2} )  $

  1. $ 1.8 \times 10^{3} $ times
  2. $ 2.4 \times 10^{3} $ times
  3. $ 1.7 \times 10^{3} $ times
  4. $ 2.1 \times 10^{3} $ times
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Energy used to go up $=mgh=60\times 10\times 10=6000\,J$

Energy used to come down $\dfrac{6000}{2}=3000\,J$

Energy used in one round trip $=9000\,J$

$1\,cal=4.5\,J$

$1\,J=\dfrac{1}{4.2\,cal}$

$9000\,J=\dfrac{9000}{4.2}=2142.85\,cal$

$7000\,kilo\,cal$ is required to burn $1\,kg$ mass

To reduce $5\,kg$ mass, energy required $=7000\times 5=35000\,kilo\,val$

Number of trip $=\dfrac{35000\times 1000}{2142.85}=1.7\times 10^{3}$
Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A steel ball of mass $5$ ${ g }$ is thrown downward with velocity $10$ ${ ms } ^ { - 1 }$ from height $19.5$ ${ m }$ . It penetrates sand by $50$ ${ cm }$ . The change in mechanical energy will be ( ${ g } = 10$ ${ ms } ^ { - 2 }$ )

  1. $1$ ${J}$
  2. $1.25$ ${J}$
  3. $1.5$ ${J}$
  4. $1.75$ ${J}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} The\, \, change\, \, in\, \, mechanic\, \, energy\, \, \Delta U=mg\left( { h+x } \right) +\frac { 1 }{ 2 } m{ v^{ 2 } } \ here\, \, m=5g=0.00\, 5kg\cdot h=19.5\, mx=50cm=0.5m,v=10\, m/s \ So,\, \Delta U=0.005\times 10919.5+0.5+\frac { 1 }{ 2 } \times 0.005\times { \left( { 10 } \right) ^{ 2 } }=0.005\times 10\times 20+\frac { 1 }{ 2 } \times 0.005\times 100=1.25J \end{array}$

Hence,
option $(B)$ is correct answer.