Physics

Thermal Properties and Thermodynamics

431 Questions

Thermal properties and thermodynamics questions evaluate concepts of heat transfer, thermal efficiency, and temperature variations. Problems involve calculating heat content, conductivity, and the performance of heat engines. This subject is regularly tested in physics sections across multiple competitive platforms.

Heat transfer calculationsThermal efficiencyBlack body radiationTemperature variationsRefrigeration performance

Thermal Properties and Thermodynamics Questions

Multiple choice introduction to ratio and percentages comparing quantities maths

The temperature of a metal coin is increased ny $100^0$C and its diameter by 0.15%. Its area increases by nearly

  1. 0.15%

  2. 0.60%

  3. 0.30%

  4. 0.0225%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$A = \pi r^2$
$\displaystyle \frac{\Delta A}{A} = 2 \frac{\Delta A}{r}$
$\displaystyle \frac{\Delta A}{A}$% $= 2 \displaystyle \left ( \frac{\Delta A}{r} \right ) \times 100$
$\displaystyle \frac{\Delta A}{A}$% $= 2 \times 0.15 = 0.30$%

Multiple choice physics reflection of light at curved surfaces uses of curved mirrors uses of spherical mirror uses of spherical mirrors

The temperature in a spherical reflector type solar cooker is raised to more than $500^o C$ by :

  1. concentrating energy at one point by using a concave reflector

  2. proper insulation of the cooker

  3. concentrating energy at one point by using a covex reflector

  4. both (A) and (B)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The temperature in a spherical reflector type solar cooler is raided to more than $500^\circ C $ by concentrating energy at one point by using a concave reflector.

Multiple choice physics heat - measurement application of various thermometric scales different types of thermometers measuring temperature introduction to temperature

Consider two thermometers $T _1$ and $T _2$ of equal length which can be used to measure temperature over the range $\theta _1$ and $\theta _2$. $T _1$ contains mercury as thermometric liquid while $T _2$ contains bromine. The volumes of the two liquids are the same at the temperature $\theta _1$. The volumetric coefficients of expansion of mercury and bromine are $18\times 10^{-5}K^{-1}$ and $108\times 10^{-5}K^{-1}$, respectively. The increase in length of each liquid is the same for the same increase in temperature. If the diameters of the capillary tubes if the two thermometers are $d _1$ and $d _2$ respectively, then the ratio $d _1:d _2$ would be closest to.

  1. $6.0$
  2. $2.5$
  3. $0.5$
  4. $0.4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Increase in length of each liquid is same 

$\dfrac{\Delta V _{hg}}{\pi d _1^2}=\dfrac{\Delta V _{br}}{\pi d _2^2}$
$\dfrac{\Delta V _{hg}\Delta\theta}{\pi d _1^2}=\dfrac{\Delta V _{br}\Delta\theta}{\pi d _2^2}$
$\dfrac{d _12}{d _2^2}=\dfrac{\gamma _{hg}}{\gamma _{br}}=\dfrac{1}{6}$
$\dfrac{d _1}{d _2}=0.4$

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

Work done by restoring force in a string within elastic limit is $-10\ J$. The maximum amount of heat produced in the string is :

  1. $10\ J$
  2. $20\ J$
  3. $5\ J$
  4. $15\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The work done by a restoring force is negative when the string is stretched. The energy stored in the string is converted into heat when the string returns to its equilibrium position, equal in magnitude to the work done.

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

An ideal heat engine has an efficiency $ \eta$ . The co-efficient of performance of the engine when driven backward is 

  1. $1-\dfrac{1}{\eta}$
  2. $\dfrac{\eta}{1- \eta}$
  3. $\dfrac{1}{\eta}-1$
  4. $\dfrac{1}{1- \eta}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Efficiency n = W/Q1 = (Q1-Q2)/Q1. Coefficient of performance (COP) for a refrigerator is Q2/W. Since W = Q1 - Q2, COP = Q2/(Q1-Q2). Dividing by Q1 gives (Q2/Q1) / (1 - Q2/Q1). Since n = 1 - Q2/Q1, Q2/Q1 = 1 - n. Substituting gives (1-n)/n = (1/n) - 1.

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

The statement "It is impossible  to construct a heat engine which can convert heat directly to work completely" was given by

  1. Clausius

  2. Carnot

  3. Plank

  4. Kelvin & Plank
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The second law of thermodynamics states , that it is impossible to construct a heat engine , operating in cycle , which extracts heat and can convert it all to useful work .In other words , it is impossible for a heat engine to convert heat completely in work .This statement is given by Kelvin & Plank .

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

A heat engine absorbs $Q _1$ heat from hot reservoir and work produced by engine is $W$, then:

  1. $Q _1$ is always $= W$
  2. only in some special cases $Q _1 = W$ otherwise $Q _1$ is greater than $W$
  3. $Q _1$ is always less than $W$
  4. $Q _1$ is always greater than $W$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a heat engine , if $Q _{1}$ is the heat absorbed from reservoir , $W$ be the work done by engine and let $Q _{2}$ be the heat given to sink , then by the first law of thermodynamics 

       $Q _{1}-Q _{2}=W$    (internal energy$dU=0$ for cyclic process)
this equation gives that $Q _{1}$ is always greater than $W$ .

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

Which of the following is correct for the efficiency of a heat engine:

  1. $\eta=\dfrac{W}{Q _1}$
  2. $\eta=\dfrac{W}{Q _2-Q _1}$
  3. $\eta=\dfrac{W}{Q _2}$
  4. $\eta=\dfrac{Q _2}{Q _1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Efficiency of a heat engine is defined as the ratio of net work done per cycle by the working substance ($W$) to the heat absorbed per cycle from the source ($Q _{1}$) ,

          $\eta=\frac{W}{Q _{1}}$ 

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

Choose the correct options for the following statements :

A) First law of thermodynamics specifies the conditions under which a body can use its heat energy to produce the work.
B) Second law of thermodynamics states that heat always flows from hot body to cold body by itself.

  1. Both A and B are true

  2. Both A and B are false

  3. A is true but B is false

  4. A is false B is true
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

First law of thermodynamics just states that $dQ = dW + dU $. 


It does $not$ tell us whether the process is feasible or not neither does it provide any condition to under which the body uses any energy.
Hence, A is false.

Second law of thermodynamics states that heat always flows from hot body to cold body by itself and for heat to flow from a cold body to hot body, external work should be done. 
Thus, B is true.

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

A household refrigerator with a coefficient of performance $1.2$ removes heat from the refrigerated space at the rate of $60kJ/min$. What would be cost of running this fridge for one month (30 days) (assuming each day it is used for $4$ hours and cost of one electrical unit is $6$ Rs.)

  1. $180$ Rs.
  2. $300$ Rs.
  3. $480$ Rs.
  4. $600$ Rs.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Heat removed = 60 kJ/min. Work input = Heat / COP = 60 / 1.2 = 50 kJ/min. Total time = 30 days * 4 hours/day * 60 min/hour = 7200 min. Total work = 50 kJ/min * 7200 min = 360,000 kJ. 1 unit = 1 kWh = 3600 kJ. Units = 360,000 / 3600 = 100 units. Cost = 100 units * 6 Rs/unit = 600 Rs.

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

An inventor claims to have developed an engine that takes in $1000\  J$ of heat and produces $1500\  J$ of work during each cycle. Comment on the validity of this claim.

  1. The proposed engine claims to produce more work in a cyclic process than the amount of heat that is supplied, so it is in violation of the first law of thermodynamics.

  2. The statement is completely valid as it increases the efficiency by increasing the work output.

  3. The proposed engine claims to produce more work in a cyclic process than the amount of heat that is supplied, so it is in violation of the zeroth law of thermodynamics.

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An inventor claims to have developed an engine that takes in 1000 J of heat and produces 1500 J of work during each cycle.The proposed engine claims to produce more work in a cyclic process than the amount of heat that is supplied, so it is in violation of the first law of thermodynamics.

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed. The first law is often formulated.
${\displaystyle \Delta U=Q-W.}$

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

A heat engine takes in $700\  J$ of heat from high-temperaturere reservoir and rejects $500\  J$ of heat to a lower temperature reservoir. How much work does the engine do in each cycle?

  1. $100\ J$
  2. $20\ J$
  3. $200\ J$
  4. $10\ J$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $Q _h=700 J$ and $Q _c=500 J$

The work done by a heat engine is given by $W=Q _h=Q _c=700-500=200 J$

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

Calculate the efficiency of a Carnot engine operating between temperatures of 900 K and 300 K.

  1. $87$ %
  2. $67$ %
  3. $100$ %
  4. $45$ %
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, the temperature of hot reservoir, $T _h=900 K$ and temperature of cold reservoir, $T _c=300 K$

Thus, the efficiency of the Carnot engine is $\eta=(1-T _c/T _h)\times 100=(1-300/900)\times 100=66.67 $ % $\sim 67$ %