Heat and Mass Transfer Metals are good conductors of heat because Their atoms collide frequently They contain free electrons Their atoms are relatively far apart They have high density Their atoms collide frequently They contain free electrons Their atoms are relatively far apart They have high density ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer The energy distribution of an ideal reflector at higher temperatures is largely in the range of Wavelength has nothing to do with it Shorter wavelength Longer wavelength Remain same at all wavelengths Wavelength has nothing to do with it Shorter wavelength Longer wavelength Remain same at all wavelengths ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer A perfect black body is one which Transmits all heat radiations Reflects all heat Absorbs heat radiations of all wave lengths falling on it Is black in colour Transmits all heat radiations Reflects all heat Absorbs heat radiations of all wave lengths falling on it Is black in colour ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer Fourier's law of heat conduction is (where Q = Amount of heat flow through the body in unit time, A = Surface area of heat flow, taken at right angles to the direction of heat flow, dT = Temperature difference on the two faces of the body, dx = Thickness of the body, through which the heat flows, taken along the direction of heat flow, and k = Thermal conductivity of the body) k. (dx/dT) k. A. (dT/dx) k. (dT/dx) k. A. (dx/dT) k. (dx/dT) k. A. (dT/dx) k. (dT/dx) k. A. (dx/dT) ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer The ratio of the emissive power and absorptive power of all bodies is the same and is equal to the emissive power of a perfectly black body. This statement is known as Stefan's law Planck's law Wien's law Kirchhoff's law Stefan's law Planck's law Wien's law Kirchhoff's law ANSWER DOWNLOAD EXAMIANS APP
Heat and Mass Transfer The logarithmic mean temperature difference (tm) is given by (where Δt1 and Δt2 are temperature differences between the hot and cold fluids at entrance and exit) tm = (Δt1 - Δt2)/ loge (Δt1/Δt2) tm = loge (Δt1/Δt2)/ (Δt1 - Δt2) tm = tm = (Δt1 - Δt2) loge (Δt1/Δt2) tm = loge (Δt1 - Δt2)/ Δt1/Δt2 tm = (Δt1 - Δt2)/ loge (Δt1/Δt2) tm = loge (Δt1/Δt2)/ (Δt1 - Δt2) tm = tm = (Δt1 - Δt2) loge (Δt1/Δt2) tm = loge (Δt1 - Δt2)/ Δt1/Δt2 ANSWER DOWNLOAD EXAMIANS APP
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