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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 falls to -10^ ^ ^...

During a winter day, the atmospheric temperature falls to -10^{}^{{}^{\circ}} C, and ice begins to form on the surface of a lake. Determine the total time required for the formation of 10 cm10 \text{ cm} of ice. Assume the temperature of the entire body of water reaches 0^{}^{{}^{\circ}} C prior to the onset of ice formation. Take the density of water as 1000 kg m−31000 \text{ kg m}^{-3}, the latent heat of fusion of ice as 3.36×105 J kg−13.36 \times 10^5 \text{ J kg}^{-1}, and the thermal conductivity of ice as 1.7 \text{ W m}^{-1} ^{}^{{}^{\circ}} C^{-1}. Neglect the expansion of water upon freezing.

Options

  1. A27.5 hours27.5 \text{ hours}
  2. B54.9 hours54.9 \text{ hours}
  3. C13.7 hours13.7 \text{ hours}
  4. D2.75 hours2.75 \text{ hours}

Correct answer

A. 27.5 hours27.5 \text{ hours}

Step-by-step solution

Let xx be the thickness of ice at time tt. The rate of heat transfer through the ice layer is given by:

dQdt=KAΔTx\dfrac{dQ}{dt} = \dfrac{KA \Delta T}{x}

The heat released when a small thickness dxdx of ice forms is:

dQ=dm⋅L=(ρAdx)LdQ = dm \cdot L = (\rho A dx) L

Equating the two expressions for the rate of heat transfer:

ρALdxdt=KAΔTx\rho A L \dfrac{dx}{dt} = \dfrac{KA \Delta T}{x}

∫0xxdx=∫0tKΔTρLdt\int_{0}^{x} x dx = \int_{0}^{t} \dfrac{K \Delta T}{\rho L} dt

x22=KΔTρLt\dfrac{x^2}{2} = \dfrac{K \Delta T}{\rho L} t

t=ρLx22KΔTt = \dfrac{\rho L x^2}{2 K \Delta T}

Substituting the given values (ρ=1000 kg m−3\rho = 1000 \text{ kg m}^{-3}, L=3.36×105 J kg−1L = 3.36 \times 10^5 \text{ J kg}^{-1}, x=0.1 mx = 0.1 \text{ m}, K = 1.7 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}, \Delta T = 10^{}^{{}^{\circ}}\text{C}):

t=1000×3.36×105×(0.1)22×1.7×10t = \dfrac{1000 \times 3.36 \times 10^5 \times (0.1)^2}{2 \times 1.7 \times 10}

t=3.36×10634 st = \dfrac{3.36 \times 10^6}{34} \text{ s}

t≈98823.5 st \approx 98823.5 \text{ s}

Converting the time into hours:

t=98823.53600 hours≈27.45 hours≈27.5 hourst = \dfrac{98823.5}{3600} \text{ hours} \approx 27.45 \text{ hours} \approx 27.5 \text{ hours}
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