Water at 50^ ^ ^ C is contained in a closed cylindrical vessel of height 10 cm and cross sectional area 10 cm^2 . The walls of the vessel are adiabatic, while the flat parts are constructed of 1 -mm thick aluminium ( K

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Quantrex Academy · Accepted Answer

Correct answer: C. 0.035 s

Volume of water in the cylinder is given by:

V = A \times h = 10 \text{ cm}^2 \times 10 \text{ cm} = 100 \text{ cm}^3 = 10^{-4} \text{ m}^3

Mass of the water is:

m = \rho V = 1000 \text{ kg m}^{-3} \times 10^{-4} \text{ m}^3 = 0.1 \text{ kg}

The heat lost by the water to cool by

\Delta T = 1.0^{}^{{}^{\circ}} \text{C}

is:

Q = mc\Delta T = 0.1 \times 4200 \times 1.0 = 420 \text{ J}

Heat is lost only through the two flat parts (top and bottom) of the cylinder, as the walls are adiabatic. The total area for heat conduction is:

A_{\text{total}} = 2A = 2 \times 10 \text{ cm}^2 = 2 \times 10^{-3} \text{ m}^2

Assuming the temperature difference remains approximately constant at

\Delta T_{\text{cond}} = 50^{}^{{}^{\circ}} \text{C} - 20^{}^{{}^{\circ}} \text{C} = 30^{}^{{}^{\circ}} \text{C}

, the rate of heat loss is:

\dfrac{dQ}{dt} = \dfrac{K A_{\text{total}} \Delta T_{\text{cond}}}{d}

\dfrac{dQ}{dt} = \dfrac{200 \times 2 \times 10^{-3} \times 30}{10^{-3}} = 12000 \text{ J s}^{-1}

The time taken for the temperature to fall by

1.0^{}^{{}^{\circ}} \text{C}

is:

t = \dfrac{Q}{\dfrac{dQ}{dt}} = \dfrac{420}{12000} = 0.035 \text{ s}

Water at 50^ ^ ^ C is contained in a closed cylindrical vessel of...

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