Question text1. A shallow circular pond of radius freezes completely during very cold nights. The pond is in a geothermal area in which the temperature of the rock reaches at 190 m
depth (assume at the surface). The thermal conductivity of the intervening rock is 4.75 and the latent heat of fusion of ice is 334 .
a) Neglecting other heat sources, calculate the mass of ice that melts per hour due to the geothermal gradient, in to two significant figures. You may use the formula for thermal conductivity:where is the thermal conductivity, is the amount of heat transferred per unit time, is the distance between two isothermal planes, is the area of the surface through which the heat flows, and is the temperature difference between the two isothermal planesMelted mass of water per hour = Answer 1 Question 4[input] kg/hour [5] b) Assuming that a mass m=27kg of water is melted per hour in the pond, calculate the thickness of the melted layer at the bottom of the pond after 10 hours. Assume a density of water equal to and that the pond is cylindrical. Give the answer in millimeters to one significant figure. thickness of melted layer = Answer 2 Question 4[input] mm [5]

Question

Question text1. A shallow circular pond of radius freezes completely during very cold nights. The pond is in a geothermal area in which the temperature of the rock reaches at 190 m
depth (assume at the surface). The thermal conductivity of the intervening rock is 4.75 and the latent heat of fusion of ice is 334 .
a) Neglecting other heat sources, calculate the mass of ice that melts per hour due to the geothermal gradient, in to two significant figures. You may use the formula for thermal conductivity:where is the thermal conductivity, is the amount of heat transferred per unit time, is the distance between two isothermal planes, is the area of the surface through which the heat flows, and is the temperature difference between the two isothermal planesMelted mass of water per hour = Answer 1 Question 4[input] kg/hour [5] b) Assuming that a mass m=27kg of water is melted per hour in the pond, calculate the thickness of the melted layer at the bottom of the pond after 10 hours. Assume a density of water equal to and that the pond is cylindrical. Give the answer in millimeters to one significant figure. thickness of melted layer = Answer 2 Question 4[input] mm [5]

BlackTom AI · Accepted Answer

Fourier conduction: Q = K A ΔT / d, so use pond area A = πr^2 and convert power to mass melted with latent heat.

1) Relevant concepts
- Steady conductive heat flux through rock: K = Q d / (A ΔT) (rearranged to Q = K A ΔT / d).
- Energy to melt mass m of ice: E = m L (L = 334 kJ/kg = 334000 J/kg).
- Power Q (W) multiplied by time (s) gives energy available for melting.

2) Step-by-step calculation (part a)
- Radius r = 34 m ⇒ area A = π r^2 = π (34^2) = π (1156) = 3628.32 m^2.
- Thermal conductivity K = 4.75 W m^-1 °C^-1, ΔT = 44 °C (44 − 0), d = 190 m.
- Compute conductive power Q:
  Q = K A ΔT / d
    = 4.75 × 3628.32 × 44 / 190
    = (4.75 × 3628.32) × 44 / 190
    = 17234.51 × 44 / 190
    = 758318.57 / 190
    = 3991.15 W (J/s).
- Energy per hour = Q × 3600 s = 3991.15 × 3600 = 14,368,146.8 J.
- Mass melted per hour m = energy / L = 14,368,146.8 / 334000 = 43.018 kg.
- Rounded to two significant figures: 43 kg/hour. This matches the provided answer.

3) Step-by-step calculation (part b)
- Given melt rate m = 27 kg/hour (problem statement for part b). Over 10 hours total mass M = 27 × 10 = 270 kg.
- Volume melted V = M / ρ = 270 kg / 1000 kg·m^-3 = 0.27 m^3.
- Spread over pond area A = 3628.32 m^2 gives thickness h = V / A = 0.27 / 3628.32 = 7.44 × 10^-5 m.
- Convert to millimetres: h = 7.44 × 10^-5 m × 1000 mm/m = 0.0744 mm.
- Round to one significant figure: 0.07 mm. This matches the provided answer.

4) Why the correct answers are correct
- Part (a) directly follows from Fourier’s law and latent heat conversion; all unit conversions were handled (kJ→J, seconds/hour).
- Part (b) uses basic volume = mass/density and thickness = volume/area; the very small thickness is consistent with a large pond area.

5) Common incorrect approaches
- Omitting factor 3600 s when converting power (W) to energy per hour will undercount melted mass by 3600×.
- Using L in kJ/kg but not converting to J/kg leads to a 1000× error.
- Using diameter instead of radius or miscomputing area changes thickness dramatically.

6) Conclusion
Part (a) yields 43 kg/hour (two significant figures) and part (b) yields 0.07 mm (one significant figure), confirming the provided answers.

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