This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Conduction, Convection and Radiation”.
1. The sun shines on a 150 m2 road surface so it is at 45°C. Below the 5cm thick asphalt(average conductivity of 0.06 W/m K), is a layer of rubbles at 15°C. Find the rate of heat transfer to the rubbles.
a) 5300 W
b) 5400 W
c) 5500 W
d) 5600 W
Explanation: There is conduction through the asphalt layer.
heat transfer rate = k A ∆T/∆x = 0.06 × 150 ×(45-15)/0.05
= 5400 W.
2. A pot of steel(conductivity 50 W/m K), with a 5 mm thick bottom is filled with liquid water at 15°C. The pot has a radius of 10 cm and is now placed on a stove that delivers 250 W as heat transfer. Find the temperature on the outer pot bottom surface assuming the inner surface to be at 15°C.
Explanation: Steady conduction, Q = k A ∆T/∆x ⇒ ∆Τ = Q ∆x / kΑ
∆T = 250 × 0.005/(50 × π/4 × 0.22) = 0.796
T = 15 + 0.796 = 15.8°C.
3. A water-heater is covered with insulation boards over a total surface area of 3 m2. The inside board surface is at 75°C and the outside being at 20°C and the conductivity of material being 0.08 W/m K. Find the thickness of board to limit the heat transfer loss to 200 W ?
a) 0.036 m
b) 0.046 m
c) 0.056 m
d) 0.066 m
Explanation: Steady state conduction through board.
Q = k A ∆T/∆x ⇒ ∆Τ = Q ∆x / kΑ
∆x = 0.08 × 3 ×(75 − 20)/200 = 0.066 m.
4. On a winter day with atmospheric air at −15°C, the outside front wind-shield of a car has surface temperature of +2°C, maintained by blowing hot air on the inside surface. If the wind-shield is 0.5 m2 and the outside convection coefficient is 250 W/Km2, find the rate of energy loss through front wind-shield.
a) 125 W
b) 1125 W
c) 2125 W
d) 3125 W
Explanation: Q (conv) = h A ∆Τ = 250 × 0.5 × [2 − ( −15)] = 250 × 0.5 × 17 = 2125 W.
5. A large heat exchanger transfers a total of 100 MW. Assume the wall separating steam and seawater is 4 mm of steel, conductivity 15 W/m K and that a maximum of 5°C difference between the two fluids is allowed. Find the required minimum area for the heat transfer.
a) 180 m2
b) 280 m2
c) 380 m2
d) 480 m2
Explanation: Steady conduction
Q = k A ∆T/∆x ⇒ Α = Q ∆x / k∆Τ
A = 100 × 10^6 × 0.004 / (15 × 5) = 480 m2.
6. The black grille on the back of a refrigerator has a surface temperature of 35°C with a surface area of 1 m2. Heat transfer to the room air at 20°C takes place with convective heat transfer coefficient of 15 W/Km^2. How much energy is removed during 15 minutes of operation?
a) 202.5 kJ
b) 212.5 kJ
c) 222.5 kJ
d) 232.5 kJ
Explanation: Q = hA ∆T ∆t, Q = 15 × 1 × (35-20)×15×60 = 202500 J = 202.5 kJ.
7. A small light bulb (25 W) inside a refrigerator is kept on and 50 W of energy from the outside seeps into the refrigerated space. How much of temperature difference to the ambient(at 20°C) must the refrigerator have in its heat exchanger having an area of 1 m2 and heat transfer coefficient of 15 W/Km2 to reject the leak of energy.
Explanation: Total energy that goes out = 50+25 = 75 W
75 = hA∆T = 15 × 1 × ∆T hence ∆T = 5°C.
8. As the car slows down, the brake shoe and steel drum continuously absorbs 25 W. Assume a total outside surface area of 0.1 m2 with a convective heat transfer coefficient of 10 W/Km2 to the air at 20°C. How hot does the outside brake and drum surface become when steady conditions are reached?
Explanation: ∆Τ = heat / hA hence ∆T = [ Τ(BRAKE) − 20 ] = 25/(10 × 0.1) = 25°C
Τ(BRAKE) = 20 + 25 = 45°C.
9. A burning wood in the fireplace has a surface temperature of 450°C. Assume the emissivity to be 1 and find the radiant emission of energy per unit area.
a) 15.5 kW/m2
b) 16.5 kW/m2
c) 17.5 kW/m2
d) 18.5 kW/m2
Explanation: Q /A = 1 × σ T^4
= 5.67 × 10–8×( 273.15 + 450)4
= 15505 W/m2 = 15.5 kW/m2.
10. A radiant heat lamp is a rod, 0.5 m long, 0.5 cm in diameter, through which 400 W of electric energy is deposited. Assume the surface emissivity to be 0.9 and neglecting incoming radiation, find the rod surface temperature?
Explanation: Outgoing power equals electric power
T4= electric energy / εσA
= 400 / (0.9 × 5.67 ×10-8× 0.5 × π × 0.005)
= 9.9803 ×10^11 K4 ⇒ T = 1000K.
11. A water-heater is covered up with insulation boards over a total surface area of 3 m2. The inside board surface is at 75°C and the outside surface is at 20°C and the board material has a conductivity of 0.08 W/m K. How thick a board should it be to limit the heat transfer loss to 200 W ?
a) 0.066 m
b) 0.166 m
c) 0.266 m
d) 0.366 m
Explanation: Steady state conduction through a single layer board.
Δx = kA(ΔT)/Q
Δx = (0.08*3)*(75-20)/200 = 0.066 m.
12. Find the rate of conduction heat transfer through a 1.5 cm thick hardwood board, k = 0.16 W/m K, with a temperature difference between the two sides of 20°C.
a) 113 W/m2
b) 213 W/m2
c) 230 W/m2
d) 312 W/m2
Explanation: . q = .Q/A = k ΔT/Δx = 0.16 Wm /K × 20K/0.015 m = 213 W/m2.
13. A 2 m2 window has a surface temperature of 15°C and the outside wind is blowing air at 2°C across it with a convection heat transfer coefficient of h = 125 W/m2K. What is the total heat transfer loss?
a) 2350 W
b) 1250 W
c) 2250 W
d) 3250 W
Explanation: .Q = h A ΔT = 125 W/m2K × 2 m2 × (15 – 2) K = 3250 W.
14. A radiant heating lamp has a surface temperature of 1000 K with ε = 0.8. How large a surface area is needed to provide 250 W of radiation heat transfer?
a) 0.0035 m2
b) 0.0045 m2
c) 0.0055 m2
d) 0.0065 m2
Explanation: .Q = εσAT^4
A = .Q/(εσT4) = 250/(0.8 × 5.67 × 10-8 × 10004)
= 0.0055 m2.
Sanfoundry Global Education & Learning Series – Thermodynamics.
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