Tag: energy transformations and energy transfers

Questions Related to energy transformations and energy transfers

In the foregoing question, the amount of energy given to the system from A to B is:

  1. 50J

  2. 60J

  3. 30J

  4. 10J


Correct Option: A

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$\mathrm { kJ }$


Correct Option: A

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. $1.2\times 10^{4}J$


Correct Option: D

Work done by 100 calorie of heat is __________.

  1. 418.4 J

  2. 4.184 J

  3. 41.84 J

  4. None of these


Correct Option: A
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


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


Correct Option: C
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 $=\dfrac{35000\times 1000}{2142.85}=1.7\times 10^{3}$

What is $\Delta E$  for a system that does 5 kcal of work by the surroundings when 3 kcal heat is absorbed by the system.  

  1. -2 kcal

  2. +2 kcal

  3. +8kcal

    • 7 kcal

Correct Option: C

A bullet of mass 10 gm moving with a speed of 20 m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets stuck in it. The amount of ice that melts, if 50% of the lost kinetic energy goes to ice, will be 

  1. $0.003 g$

  2. $0.30 g$

  3. $0.0003 g$

  4. $3.0 g$


Correct Option: A

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. $1.75$ ${J}$


Correct Option: B
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.

A block of ice falls from certain height and completely melts. If only 3/4th of the energy is absorbed by the block the height of fall should be

  1. 48.4 m

  2. 84.4m

  3. 88.4m

  4. 44.8m


Correct Option: D

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


Correct Option: A