Let us assume two walls of same thickness and cross-sectional area having thermal conductivities in the ratio 1/2. Let us say there is same temperature difference across the wall faces, the ratio of heat flow will be









Answer: (b).1/2

Interact with the Community - Share Your Thoughts

Uncertain About the Answer? Seek Clarification Here.

Understand the Explanation? Include it Here.

Q. Let us assume two walls of same thickness and cross-sectional area having thermal conductivities in the ratio 1/2. Let us say there is same temperature difference across the wall...

Similar Questions

Explore Relevant Multiple Choice Questions (MCQs)

Q. The interior of an oven is maintained at a temperature of 850 degree Celsius by means of a suitable control apparatus. The oven walls are 500 mm thick and are fabricated from a material of thermal conductivity 0.3 W/m degree. For an outside wall temperature of 250 degree Celsius, workout the resistance to heat flow

Q. A plane slab of thickness 60 cm is made of a material of thermal conductivity k = 17.45 W/m K. Let us assume that one side of the slab absorbs a net amount of radiant energy at the rate q = 530.5 watt/m². If the other face of the slab is at a constant temperature t₂ = 38 degree Celsius. Comment on the temperature with respect to the slab?

Q. The rate of heat transfer for a plane wall of homogenous material with constant thermal conductivity is given by

Q. In case of homogeneous plane wall, there is a linear temperature distribution given by

Q. The rate of convective heat transfer between a solid boundary and adjacent fluid is given by

Q. A homogeneous wall of area A and thickness δ has left and right hand surface temperatures of 0 degree Celsius and 40 degree Celsius. Determine the temperature at the center of the wall

Q. A rod of 3 cm diameter and 20 cm length is maintained at 100 degree Celsius at one end and 10 degree Celsius at the other end. These temperature conditions are attained when there is heat flow rate of 6 W. If cylindrical surface of the rod is completely insulated, determine the thermal conductivity of the rod material

Q. A composite wall generally consists of

Q. Three metal walls of same thickness and cross sectional area have thermal conductivities k, 2k and 3k respectively. The temperature drop across the walls (for same heat transfer) will be in the ratio

Q. A composite wall is made of two layers of thickness δ₁ and δ₂ having thermal conductivities k and 2k and equal surface area normal to the direction of heat flow. The outer surface of composite wall are at 100 degree Celsius and 200 degree Celsius. The minimum surface temperature at the junction is 150 degree Celsius. What will be the ratio of wall thickness?

Q. Let us say thermal conductivity of a wall is governed by the relation k = k0 (1 + α t). In that case the temperature at the mid-plane of the heat conducting wall would be

Q. Heat is transferred from a hot fluid to a cold one through a plane wall of thickness (δ), surface area (A) and thermal conductivity (k). The thermal resistance is

Q. A pipe carrying steam at 215.75 degree Celsius enters a room and some heat is gained by surrounding at 27.95 degree Celsius. The major effect of heat loss to surroundings will be due to

Q. “Radiation cannot be affected through vacuum or space devoid of any matter”. True or false

Q. A composite slab has two layers having thermal conductivities in the ratio of 1:2. If thickness is same for each layer then the equivalent thermal conductivity of the slab would be

Q. A composite wall of a furnace has two layers of equal thickness having thermal conductivities in the ratio 2:3. What is the ratio of temperature drop across the two layers?

Q. Typical examples of heat conduction through cylindrical tubes are not found in

Q. The rate of heat conduction through a cylindrical tube is usually expressed as

Q. Logarithmic mean area of the cylindrical tube is given as

Q. A hot fluid is being conveyed through a long pipe of 4 cm outer diameter and covered with 2 cm thick insulation. It is proposed to reduce the conduction heat loss to the surroundings to one-third of the present rate by further covering with same insulation. Calculate the additional thickness of insulation

Recommended Subjects

Are you eager to expand your knowledge beyond Heat Transfer? We've handpicked a range of related categories that you might find intriguing.

Click on the categories below to discover a wealth of MCQs and enrich your understanding of various subjects. Happy exploring!