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

Heat flows through a composite slab, as shown below. The depth of the slab is 1 m. the k values are in W/m K. the overall thermal resistance in K/W is

  1. 17.2

  2. 21.9

  3. 28.6

  4. 39.2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\hspace{3cm}\frac{1}{R_{eq}}=\frac{1}{R_2}+\frac{1}{R_3}\\ \hspace{3cm}\frac{1}{R_{eq}}=\frac{R_3+R_2}{R_2R_3}\\ \hspace{3cm}R_{eq}=\frac{R_2R_3}{R_2+R_3}=\frac{5\times12.5}{5+12.5}=3.6K/W\\ \text{Resistance $R_1$ & $R_{eq}$ are in series .So total Resistance will be}\\ \hspace{3.5cm}R=R_1+R_{eq}=25+3.6=28.6K/W$ 

Multiple choice

A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/m.K, its density 9000 kg/m3 and its specific heat 385 J/kg.K. If the heat transfer coefficient is 250 W/m2.K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be

  1. 8.7 K/s

  2. 13.9 K/s

  3. 17.3 K/s

  4. 27.7 K/s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiple choice

In two air standard cycles - one operating on the Otto and the other on the Brayton cycle - air is entropically compressed from 300 K to 450 K, heat is added to raise the temperature to 600 K in the Otto cycle and to 550 K in the Brayton cycle. If $\eta_0$ and $\eta_B$are the efficiencies of the Otto and Brayton cycles, then
  1. $\eta_0=0.25,\eta_B=0.18$
  2. $\eta_0=\eta_B=0.33$
  3. $\eta_0=0.25,\eta_B=0.45$

  4. it is not possible to calculate the efficiencies unless the temperature after the expansion is given.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice

In two air standard cycles - one operating on the Otto and the other on the Brayton cycle - air is entropically compressed from 300 to 450 K, heat is added to raise the temperature to 600 K in the Otto cycle and to 550 K in the Brayton cycle. If Wo and WB are work outputs per unit mass, then

  1. Wo >WB

  2. Wo <WB

  3. Wo =WB

  4. It is not possible to calculate the work outputs unless the temperature after the expansion is given.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Multiple choice

An un-insulated air conditioning duct of rectangular cross section 1 m × 0.5 m, carrying air at 20°C with a velocity of 10 m/s, is exposed to an ambient of 30°C. Neglect the effect of duct construction material. For air in the range of 20°C-30°C, data are as follows: thermal conductivity =0.025 W/m.K; viscosity = 18$\mu$ Pa.s; Prandtl number = 0.73; density = 1.2 kg/m3. The laminar flow Nusselt number is 3.4 for constant wall temperature conditions and, for turbulent flow, Nu=0.023 Re0.8 Pr0.33. The Reynolds number for the flow is

  1. 444

  2. 890

  3. 4.44 × 105

  4. 5.33 × 105
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiple choice

An un-insulated air conditioning duct of rectangular cross section 1 m × 0.5 m, carrying air at 20°C with a velocity of 10 m/s, is exposed to an ambient of 30°C. Neglect the effect of duct construction material. For air in the range of 20°C-30°C, data are as follows: thermal conductivity =0.025 W/m.K; viscosity = 18$\mu$ Pa.s; Prandtl number = 0.73; density = 1.2 kg/m3. The laminar flow Nusselt number is 3.4 for constant wall temperature conditions and, for turbulent flow, Nu = 0.023 Re0.8 Pr0.33. The heat transfer per metre length of the duct is

  1. 3.8 <span style="font-family:" arial;="" font-size:="" small;="">watt

  2. 5.3 <span style="font-family:" arial;="" font-size:="" small;="">watt

  3. 89 <span style="font-family:" arial;="" font-size:="" small;="">watt

  4. 769 <span style="font-family:" arial;="" font-size:="" small;="">watt
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\frac{O}{L}=25.64\times3\times10=769.2W\simeq769W$

Multiple choice

Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200$\Omega$, the current passing through the electrodes is approximately

  1. 1480 A

  2. 3300 A

  3. 4060 A

  4. 9400 A
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiple choice physics electric current and its effects coil and fuse heating effect of electric current- uses applications of heating effect of electric current

Thermal energy produced by the resistance R in time duration 1 $ \mu s $. using the source at resonance condition is 

  1. 0 J
  2. $ 1 \mu J $
  3. $ 100 \mu J $

  4. not possible to calculate from the given information.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

To calculate thermal energy produced by resistance, we need the current through the resistance, which requires knowledge of the circuit components (voltage source, impedance at resonance, circuit configuration). The question states 'using the source at resonance condition' but provides no numerical values for voltage, current, resistance, frequency, or circuit parameters. Without this information, the energy cannot be calculated.

Multiple choice physics electric current and its effects coil and fuse heating effect of electric current- uses applications of heating effect of electric current

If p.d. across a conductor is constant and resistivity of its material is $\rho $, the joule heat produced in 1 s is proportional to .. 

  1. $\dfrac{1}{\rho }$
  2. $\dfrac{1}{{\sqrt \rho }}$
  3. $\rho $
  4. ${{\rho ^2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Joule heat in 1 second is H = V²/R (since P.d. is constant). Resistance of a conductor is R = ρL/A, where L is length and A is cross-sectional area. For a fixed geometry (L and A constant), R ∝ ρ. Therefore H = V²/R ∝ 1/R ∝ 1/ρ. The heat produced is inversely proportional to resistivity.