Tag: energy transformations and energy transfers

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Questions Related to energy transformations and energy transfers

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.

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

A block of ice at 0 C whose mass is initially 50.0 kg slides along a horizontal surface starting at a speed of 5.38 m/s and finally coming of ice melted as a result of the friction between the block and the surface will be

  1. 2.16 g

  2. 4.0 g

  3. 1 g

  4. 50 g
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial KE = 0.5 * 50 * 5.38^2 = 723.6 J. Heat = 723.6 J. Mass melted = Heat / Latent heat = 723.6 / 334,000 J/kg = 0.00216 kg = 2.16 g.