Question

Water at 50°C is filled in a closed cylindrical vessel of height 10 cm and cross sectional area 10 cm2. The walls of the vessel are adiabatic but the flat parts are made of 1-mm thick aluminium (K = 200 J s−1 m−1°C−1). Assume that the outside temperature is 20°C. The density of water is 100 kg m−3, and the specific heat capacity of water = 4200 J k−1 g °C−1. Estimate the time taken for the temperature of fall by 1.0 °C. Make any simplifying assumptions you need but specify them.

Answer 1

Answer:

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

Answer 2

Answer:

$t=0.7\ s$ to raise the temperature by 1°C

Explanation:

Given:

• initial temperature of water, $T_i=50^{\circ}C$

• area subjected to aluminium, $A=0.001\ m^2$

• thermal conductivity of aluminium, $k=200\ W.m^{-1}.^{\circ}C^{-1}$

• thickness of aluminium, $dx=0.001\ m$

• outside temperature of the aluminium sheet, $T_o=20^{\circ}C$

• specific heat capacity of water, $c=4200\ J.kg^{-1}.K^{-1}$

Amount of heat transfer that occurs for 1°C change in temperature.

$Q=4200\ J$

Now,  using Fourier's law of conduction we find the rate of heat transfer:

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=200\times 0.001\times \frac{50-20}{0.001}$

$\dot Q=6000\ W$

∴Time required to raise the temperature (i.e. transfer the heat of 4200 J):

$t=\frac{Q}{\dot Q}$

$t=\frac{4200}{6000}$

$t=0.7\ s$

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TOPIC: Fourier's law of thermal conduction

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