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
View Answer
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.
a) 15.8°C
b) 16.8°C
c) 18.8°C
d) 19.8°C
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
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.
a) 0°C
b) 5°C
c) 10°C
d) 15°C
View Answer
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?
a) 25°C
b) 35°C
c) 45°C
d) 55°C
View Answer
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
View Answer
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?
a) 700K
b) 800K
c) 900K
d) 1000K
View Answer
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
View Answer
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
View Answer
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
View Answer
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
View Answer
Explanation: .Q = εσAT^4
A = .Q/(εσT4) = 250/(0.8 × 5.67 × 10-8 × 10004)
= 0.0055 m2.
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