Answer:
9680 cal
Explanation:
A = cross-sectional area of the bar = 30 cm² = 30 x 10⁻⁴ m²
L = length of the bar = 1.5 m
T₁ = Temperature at one end of the bar = 25 °C
T₂ = Temperature at other end of the bar = 300 °C
k = Thermal conductivity of Aluminum = 1.76 x 10⁴ Cal cm /(m² ⁰C)
Q = amount of heat conducted per second
Amount of heat conducted per second is given as
[tex]Q = \frac{k A (T_{2} - T_{1})}{L}[/tex]
Q = (1.76 x 10⁴) (30 x 10⁻⁴) (300 - 25)/1.5
Q = 9680 cal
