A uniform slab measuring 10 cm 10 cm 1 cm is placed between two heat reservoirs maintained at temperatures of 10^ C and 90^ C . The larger surface areas of the slab are in contact with the reservoirs. If the thermal conductivity of the material is 0.80 W m ⁻¹ ^ C ⁻¹ , determine the amount of heat flowing through the slab per minute.

Step-by-step solution

Area of the larger surface, A = 10 cm 10 cm = 100 cm ^2 = 10⁻² m ^2 Thickness of the slab, d = 1 cm = 10⁻² m Temperature difference, T = 90^ C - 10^ C = 80^ C Time, t = 1 minute = 60 s Thermal conductivity, K = 0.80 W m ⁻¹ ^ C ⁻¹ The amount of heat flowing through the slab is given by Fourier's law of heat conduction: Q = K A T t d Substituting the given values: Q = 0.80 10⁻² 80 60 10⁻² Q = 0.80 80 60 Q = 3840 J