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# A Steel Frame (K = 45 W M−1°C−1) of Total Length 60 Cm and Cross Sectional Area 0.20 Cm2, Forms Three Sides of a Square. the Free Ends Are Maintained at 20°C and 40°C. Find the Rate of Heat Flow - Physics

Sum

A steel frame (K = 45 W m−1°C−1) of total length 60 cm and cross sectional area 0.20 cm2, forms three sides of a square. The free ends are maintained at 20°C and 40°C. Find the rate of heat flow through a cross section of the frame.

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#### Solution

Thermal conductivity, K = 45 W m–1 °C–1

Length, l = 60 cm = 0.6 m

Area of cross section, A = 0.2 cm2 = 0.2 × 10−4 m2

Initial temperatureT1 = 40°C

Final temperatureT2 = 20°C

Rate of flow of heat = "Tempreature diffrences"/ "thermal resistance"

(DeltaQ)/(Deltat) = (KA ( T_1 - T_2 ))/ ( l )}

(DeltaQ)/(Deltat) = (45 xx 0.2 xx 10^-4 ( 40 - 20))/(0.6)

= 0.03  W

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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 6 Heat Transfer
Q 8 | Page 98
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