What is the steady state rate of heat flow through a pane of glass that is 40.0 cm by 30.0 cm with a thickness of 4.00 mm when the outside temperature of the glass is −10.0℃ and its inside temperature is 25.0℃? The thermal conductivity of glass is 0.105 W/(m⋅K); the specific heat of glass is 0.180 cal/(g⋅℃); and 1 cal = 4.190 J.

110.25 watt/s

Explanation:

The rate of heat transfer is given by the formula

R = k*A(T1 - T2) / d, where

k = thermal conductivity, 0.105 (W/(m*K)

d = thickness, 4 mm = 0.004 m

T1 = inside temperature, 25° C

T2 = outside temperatures, -10° C

A = area 0.3 * 0.4 = 0.12 m²

Now, applying the given data to the formula, we have

Rate = 0.105 * 0.12 (25 - -10) / 0.004

Rate = 0.105 * 0.12(35) / 0.004

Rate = 0.105 * 4.2 / 0.004

Rate = 0.441 / 0.004

Rate = 110.25 Watt/s

Therefore, the steady state rate of heat flow is 110.25 watt/s

v=ir = 3x15 =45 v