Conduction Law
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The temperature of surroundings is 10∘ C. [Assume Newton's Law of cooling is applicable]
- T=32∘ C
- T=28∘ C
- T=24∘ C
- T=36∘ C
- K1+K22
- 3(K1+K2)2
- K1+K2
- 2(K1+K2)
- 2.0 s
- 3.0 s
- 4.5 s
- 6.0 s
- 1600 W
- 2067 W
- 2533 W
- 3933 W
1. Steady- state one-dimensional conduction.
2. Constant value of thermal conductivity
3. The bounding surfaces are isothermal.
4. Temperature gradient is constant.
of these statements:-
- 2 and 4 are correct
- 1, 3 and 4 are correct
- 1 , 2 , and 3 are correct
- 2 , 3 , and 4 are correct
- 50∘C
- 100∘C
- 75∘C
- 125∘C
- 19.2∘C
- 16.4∘C
- 11.5∘C
- 22∘C
Why does the heat transfer coefficient increase with velocity?
- 1 minute
- 2 minutes
- 4 minutes
- 16 minutes
- [ML1T2]
- [ML1T−2]
- [ML−1T2]
- [ML−1T−2]
A uniform slab of dimension 10 cm × 10 cm × 1 cm is kept between two heat reservoirs at temperatures 10∘C and 90∘C. The larger surface areas touch the reservoirs. The thermal conductivity of the material is 0.80Wm−1∘C−1 Find the amount of heat flowing through the slab per minute.
- 16.5∘C
- 22.5∘C
- 28.5∘C
- 32.5∘C
- 15a4
- 2a3
- 12a5
- 3a
Steam at 120∘C is continuously passed through a 50 -cm long rubber tube of inner and outer radii 1.0 cm and 1.2 cm. The room temperature is 30∘C. Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber = 0.15Js−1m−1∘C−1.
In the relation P=αβ eαxnRθ, P is power, x is distance, n is number of moles, R is gas constant and θ is temperature. The dimensional formula of β is
[M0L0T0]
[M1L0T1]
[M0L−1T1]
[M0L−1T−1]
- C
- C
- C
- C
- 150 W
- 150 W
- 75 W
- 750 W
- 100∘C
- 1∘C
- 10∘C
- 1000∘C
Two plates of the same area and the same thickness having thermal conductivities k1 and k2 are placed one on top of the other. The top and bottom faces of the composite plate are maintained at different constant temperatures. The thermal conductivity of the composite plate will be
k1k2(k1+k2)
(k1+k2)
2k1k2(k1+k2)
12(k1+k2)
A pitcher with 1 -mm thick porous walls contains 10 kg of water. Water comes to its outer surface and evaporates at the rate of 0.1gs−1. The surface are aof the pitcher (one side) = 200 cm2. The room temperature = 42∘, latent heat of vaporization = 2.27×106Jkg−1, and the thermal conductivity of the porous walls = 0.80Js−1m−1∘C−1. Calculate the temperature of water in the pitcher when it attains a constant value.
Four identical rods AB, CD, CF and DE are joined as shown in figure. The length, cross -sectional area and thermal conductivity of each rod are l, A and K respectively. The ends A, E and F are maintained at temperatures T1, T2 and T3 respectively. Assuming no loss of heat to the atmosphere, find the temperature at B.
The normal body-temperature of a person is 97∘F Calculate the rate at which heat is flowing out of his body through the clothes assuming the following values. Room temperature = 47∘F, surface of the body under clothes = 1.6m2, conductivity of the cloth = 0.04Js−1m−1∘C−1, thickness of the cloth = 0.5 cm.
- r= 2 cm, l=0.5 m
- r= 1 cm, l=0.5 m
- r= 1 cm, l=1 m
- r= 2 cm, l=2 m
- Rate of heat flow is proportional to the cross-sectional area of rod
- Rate of heat flow is proportional is the temperature difference between the ends of the rod
- Rate of heat flow is inversely proportional to thickness of the rod
- Rate of heat flow is proportional to the coefficient of thermal conductivity of material of the rod
A cubical box of volume 216cm3 is made up of 0.1 cm thick wood. The inside is heated electrically by a 100 W heater. It is found that the temperature difference between the inside and the outside surface is 5∘C in steady state. Assuming that the entire electrical energy spent appears as heat, find the thermal conductivity of the material of the box.
- K1T1+K2T2d1+d2
- K1T1d2+K2T2d1K1d2+K2d1
- (K1d1+K2d2)T1T2T1+T2
- K1d1T1+K2d2T2K1d1+K2d2
- 1
- 8
- 16
- 2