Tag: heat, internal energy and work

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Questions Related to heat, internal energy and work

Multiple choice physics heat energy transfers heat energy heat, internal energy and work 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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice physics heat energy transfers heat energy heat, internal energy and work internal energy

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. $\frac{t _1 + t _2}{t _1 + 273}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The coefficient of performance (COP) for a heat pump is defined as Q_h / W. In terms of absolute temperatures (Kelvin), the heat delivered to the room for each unit of work is T_1 / (T_1 - T_2), where T_1 and T_2 are in Kelvin.

Multiple choice physics heat energy transfers heat energy heat, internal energy and work internal energy

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The heat energy required is calculated using Q = m * c * delta_T. Here, m = 5 kg, c = 4.2 kJ/kg C, and delta_T = 100 - 20 = 80 C. Thus, Q = 5 * 4.2 * 80 = 1680 kJ.

Multiple choice physics heat energy transfers heat energy heat, internal energy and work internal energy

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. $80R$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice physics heat energy transfers heat energy heat, internal energy and work internal energy

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. $\dfrac{{{n^2}{h^2}}}{{4m{L^2}}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The energy levels of a particle in a one-dimensional infinite potential well of width L are given by E_n = (n^2 * h^2) / (8 * m * L^2). This is a standard result from quantum mechanics.

Multiple choice physics heat energy transfers heat energy heat, internal energy and work internal energy

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. $52.875\ g$
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

$M _{i} =52.875 g$
Multiple choice physics heat energy transfers heat energy heat, internal energy and work internal energy

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=3.36\times 10^{5}J/kg$

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

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

$=\dfrac{1}{33}$

Multiple choice physics heat energy transfers heat energy heat, internal energy and work internal energy

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. $0.75\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$=\dfrac{1}{2}m(v^2 _1-v^2 _2)$

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

$=3.75J$