Aditi Chauhan
Last Activity: 10 Years ago
Sol. A = 10cm^2, h = 10 cm
∆Q/∆t = KA(θ base 1 - θ base 2)ℓ = 200 * 10^-3 * 30/1 * 10^-3 = 6000
Since heat goes out form both surfaces. Hence net heat coming out.
= ∆Q/∆t = 6000 * 2 = 12000, ∆Q/∆t = MS ∆θ/∆t
⇒ 6000 * 2 * 10^-3 * 10^-1 * 1000 * 4200 * ∆θ/∆t
⇒ ∆θ/∆t = 72000/420 = 28.57
So, in 1 Sec. 28.57°C is dropped
Hence for drop 1°C 1/28.57 sec. = 0.0035 sec. is required