Tag: internal energy

Questions Related to internal energy

Select the option which describes the best definition of heat in the context of thermal energy?

  1. energy possessed by a system due to movement of particles within the system

  2. the measure of thermal energy

  3. thermal energy transferred from one object to another

  4. energy possessed by an object due to the temperature of the object

  5. potential energy


Correct Option: C
Explanation:

Heat is a thermal energy that is transferred from hotter body to a cooler body.

The mode of exchange of energy due to temperature difference is known as:

  1. heat energy

  2. electric energy

  3. potential energy

  4. kinetic energy


Correct Option: A
Explanation:

Heat energy is termed as total energy contained within a substance.

The temperature inside a refrigerator is $t _2$ $^0C$ and the room temperature is $t _1$ $^0C$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be   

  1. $\frac{t _1}{t _1 - t _2}$

  2. $\frac{t _1 + 273}{t _1 - t _2}$

  3. $\frac{t _2 + 273}{t _1 + t _2}$

  4. $\frac{t _1 + t _2}{t _1 + 273}$


Correct Option: B

How much heat energy is gained when 5kg of water at $20^{ _-^0}C$ I brought to its boiling point? (Specific heat of water = 4.2 kJ $kg^{-1} C^{-1}$) -

  1. 1680 kJ

  2. 1700 kJ

  3. 1720 kJ

  4. 1740 kJ


Correct Option: A

How much heat energy in joules must be supplied to $14\ gms$ of nitrogen at room temperature to rise its temperature by $ 40^o C $ at constant pressure?
(Mol.wt.of $ N _2 = 28 gm , R = constant $)

  1. $50R$

  2. $60R$

  3. $70R$

  4. $80R$


Correct Option: C
Explanation:

Given,

Mass of nitrogen, $m=14gms$
Temperature, $T=40^0C$
Molecular weight of nitrogen, $M=28gms$
Now,
We know that,
Amount of heat supplied, $Q = n{C _p}\left( {dT} \right)$
No, of moles, $n = \frac{m}{M}$
$\therefore Q = \frac{m}{M}{C _p}dT$
$\begin{array}{l} =\frac { { 14 } }{ { 28 } } \times \frac { { 7R } }{ 2 } \times 40 \ =70R \end{array}$
Hence,
Option $C$ is correct answer.

Confined particles energy is given by 

  1. $\dfrac{{{n^2}{h^2}}}{{2m{L^2}}}$

  2. $\dfrac{{2{n^2}{h^2}}}{{m{L^2}}}$

  3. $\dfrac{{{n^2}{h^2}}}{{8m{L^2}}}$

  4. $\dfrac{{{n^2}{h^2}}}{{4m{L^2}}}$


Correct Option: C

At $3{ 0 }^{ \circ  }C$, a lead bullet of $50\ g$, is fired vertically upwards with a speed of $840\ m/s$. The specific heat of lead is $0.02\ cal/{ g }^{ \circ  }C$. on returning to the starting level, it strikes to a cake of ice at $0^{ \circ  }C$. Calculate the amount of ice melted (Assume all the energy is spent in melting only)

  1. $62.7\ g$

  2. $55\ kg$

  3. $52.875\ kg$

  4. $52.875\ g$


Correct Option: D
Explanation:
Kinetic energy of bullet = $\dfrac{mv^2}{2}$=$\dfrac12\times0.02\times840^2$ = $17640 joule =4200 cal$

heat supplied by lead to ice : $\Delta H _1 = ms\Delta T$ = $50\times 0.02 \times 30$ = $ 30 cal$

Total heat supplied = $4200 + 30 = 4230 cal$

let $M _{i}$ mass of ice melted $L$ is the latent heat of ice

$M _{i}L =4230$
$M _{i} \times 80 =4230$

$M _{i} =52.875 g$

A refrigerator has to transfer an average of $263J$ of heat per second from the temperature $-10^oC$ to $25^oC.$ The average power consumed, if no energy is lost is 

  1. $100W$

  2. $7.5W$

  3. $25W$

  4. $35W$


Correct Option: A

Hailstone at $0^{o}C$ falls from a height of $1\ km$ on an insulating surface converting whole of its kinetic energy into heat. What part of it will melt ? $(g=10\ m/s^{2}$)

  1. $\dfrac{1}{33}$

  2. $\dfrac{1}{8}$

  3. $\dfrac{1}{33}\times 10^{-4}$

  4. All of it will melt


Correct Option: A
Explanation:

Given:

The height from which the Hailstone falls is $1\ km$.

The energy lost by the hailstone is the potential energy and as it falls, this energy converts into heat energy. This heat energy is utilized in melting the hailstone.

So, the part of the hailstone melted is given by equating the potential energy to the latent heat of fusion.
$mgh=KmL$
$\because$ Latent heat of ice $=3.36\times 10^{5}J/kg$

$\Rightarrow K=\dfrac{gh}{L}$

$=\dfrac{10\times 1000}{3.36\times 10^{5}}$

$=\dfrac{1}{33}$

A block of mass $100\ g$ slides on a rough horizontal surface. If the speed of the block decreases from $10\ m/s^{-1}$ to $5\ m/s^{-1}$, the thermal energy developed in the process is:

  1. $3.75\ J$

  2. $37.5\ J$

  3. $0.375\ J$

  4. $0.75\ J$


Correct Option: A
Explanation:
Given:
The mass of the block is $100\ g$
The initial speed of the block is $10\ m/s$
The final speed of the block is $5\ m/s$

The thermal energy developed in the process is due to the lowering of the speed i.e. the reduction in the kinetic energy of the block.

Thermal energy = loss in kinetic energy
$=\dfrac{1}{2}m(v^2 _1-v^2 _2)$

$=\dfrac{1}{2}100\times 10^{-3}(10^2-5^2)$

$=3.75J$