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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Heat Transfer

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat Transfer

During a winter day, the atmospheric temperature drops to -10^ ^ ^...

During a winter day, the atmospheric temperature drops to -10^{}^{{}^{\circ}} \text{C} and ice forms on the surface of a lake. Suppose that the water temperature at the bottom of the lake stays constant at 4^{}^{{}^{\circ}} \text{C} while the ice forms (the heat required to maintain the bottom layer's temperature may be supplied by the lake bed). The lake has a depth of 1.0 m1.0 \text{ m}. Determine the steady state maximum value of the ice thickness. The thermal conductivity of water is 0.50 \text{ W m}^{-1}^{}^{{}^{\circ}}\textC^{-1}, the thermal conductivity of ice is 1.7 \text{ W m}^{-1}^{}^{{}^{\circ}}\textC^{-1}, the density of water is 1000 kg m−31000 \text{ kg m}^{-3}, and the latent heat of fusion of ice is 3.36×105 J kg−13.36 \times 10^5 \text{ J kg}^{-1}. Neglect the expansion of water upon freezing.

Options

  1. A0.89 m0.89 \text{ m}
  2. B0.42 m0.42 \text{ m}
  3. C0.50 m0.50 \text{ m}
  4. D0.77 m0.77 \text{ m}

Correct answer

A. 0.89 m0.89 \text{ m}

Step-by-step solution

Let the maximum steady-state thickness of the ice be xx. The thickness of the water layer below the ice is 1.0−x1.0 - x.

At the steady state of maximum ice thickness, the rate of heat conducted upward through the water layer equals the rate of heat conducted upward through the ice layer. The temperature at the ice-water interface is 0^{}^{{}^{\circ}} \text{C}.

The rate of heat flow per unit area through the water layer is:

QwAt=kw(Tb−Ti)1.0−x\dfrac{Q_w}{A t} = \dfrac{k_w (T_b - T_i)}{1.0 - x}

The rate of heat flow per unit area through the ice layer is:

QiAt=ki(Ti−Ta)x\dfrac{Q_i}{A t} = \dfrac{k_i (T_i - T_a)}{x}

Equating the two heat flow rates:

kw(4−0)1.0−x=ki(0−(−10))x\dfrac{k_w (4 - 0)}{1.0 - x} = \dfrac{k_i (0 - (-10))}{x}

Substituting the given values k_w = 0.50 \text{ W m}^{-1}^{}^{{}^{\circ}}\textC^{-1} and k_i = 1.7 \text{ W m}^{-1}^{}^{{}^{\circ}}\textC^{-1}:

0.50×41.0−x=1.7×10x\dfrac{0.50 \times 4}{1.0 - x} = \dfrac{1.7 \times 10}{x}

21.0−x=17x\dfrac{2}{1.0 - x} = \dfrac{17}{x}

2x=17(1.0−x)2x = 17(1.0 - x)

2x=17−17x2x = 17 - 17x

19x=1719x = 17

x=1719≈0.89 mx = \dfrac{17}{19} \approx 0.89 \text{ m}
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