Question text1. A shallow circular pond of radius [math: r=24m] freezes completely during very cold nights. The pond is in a geothermal area in which the temperature of the rock reaches [math: 45∘C] at 450 m
depth (assume [math: 0∘C] at the surface). The thermal conductivity of the intervening rock is 4.75[math: Wm−1∘C−1] and the latent heat of fusion of ice is 334 [math: kJ/kg]. a) Neglecting other heat sources, calculate the mass of ice that melts per hour due to the geothermal gradient, in [math: kg/hour] to two significant figures. You may use the formula for thermal conductivity:[math: K=QdAΔT]where [math: K] is the thermal conductivity, [math: Q] is the amount of heat transferred per unit time, [math: d] is the distance between two isothermal planes, [math: A] is the area of the surface through which the heat flows, and [math: ΔT] is the temperature difference between the two isothermal planesMelted mass of water per hour = Answer 1 Question 2[input] kg/hour [5] b) Assuming that a mass m=36kg of water is melted per hour in the pond, calculate the thickness of the melted layer at the bottom of the pond after 7 hours. Assume a density of water equal to [math: 1000kg/m3] and that the pond is cylindrical. Give the answer in millimeters to one significant figure. thickness of melted layer = Answer 2 Question 2[input] mm [5]

Question

Question text1. A shallow circular pond of radius [math: r=24m] freezes completely during very cold nights. The pond is in a geothermal area in which the temperature of the rock reaches [math: 45∘C] at 450 m
depth (assume [math: 0∘C] at the surface). The thermal conductivity of the intervening rock is 4.75[math: Wm−1∘C−1] and the latent heat of fusion of ice is 334 [math: kJ/kg]. a) Neglecting other heat sources, calculate the mass of ice that melts per hour due to the geothermal gradient, in [math: kg/hour] to two significant figures. You may use the formula for thermal conductivity:[math: K=QdAΔT]where [math: K] is the thermal conductivity, [math: Q] is the amount of heat transferred per unit time, [math: d] is the distance between two isothermal planes, [math: A] is the area of the surface through which the heat flows, and [math: ΔT] is the temperature difference between the two isothermal planesMelted mass of water per hour = Answer 1 Question 2[input] kg/hour [5] b) Assuming that a mass m=36kg of water is melted per hour in the pond, calculate the thickness of the melted layer at the bottom of the pond after 7 hours. Assume a density of water equal to [math: 1000kg/m3] and that the pond is cylindrical. Give the answer in millimeters to one significant figure. thickness of melted layer = Answer 2 Question 2[input] mm [5]

BlackTom AI · Accepted Answer

We start by restating the two parts of the problem and the data involved, then examine each proposed answer in turn.

Part a) Melt rate due to geothermal gradient
- Given: radius r = 24 m, so the surface area through which heat transfers is A = π r^2 = π × (24 m)^2 = π × 576 m^2 ≈ 1810 m^2.
- Temperature difference: ΔT = 45°C − 0°C = 45°C.
- Thermal conductivity: K = 4.75 W m−1 °C−1.
- Distance between isothermal planes: d = 450 m.
- The heat transfer rate is K A ΔT / d, i.e., Q̇ = (4.75 × 1810 × 45) / 450 W.
- Numerically, 4.75 × 1810 ≈ 8597.5; multiplying by 45 gives ≈ 386,887.5; dividing by 450 yields Q̇ ≈ 860 W (approximately).
- This is the heat input per second. To convert to energy per hour, Q̇ per hour = 860 W × 3600 s/h ≈ 3.10 × 10^6 J/h.
- The latent heat of fusion of ice is L ≈ 334 kJ/kg = 334,000 J/kg. So the mass melted per hour is ṁ = Q̇_hour / L ≈ (3.10 × 10^6 J/h) / (3.34 × 10^5 J/kg) ≈ 9.3 kg/h.
- Therefore, the physically consistent melt rate is about 9 kg/hour (to two significant figures this is 9.0 kg/h). If one were to compare to the provided candidate value 1.4×10^3 kg/h, note that 1400 kg/h would require roughly 100× more heat transfer or a much smaller melting enthalpy, which is not supported by the given data. The discrepancy indicates the 1400 kg/h figure is not aligned with the provided K, A, ΔT, and d values.

Part b) Thickness of the melted layer after 7 hours, given a melt rate m = 36 kg/h
- Melt rate is 36 kg of water per hour. The density of water is 1000 kg/m^3, so the melted water volume per hour is V̇ = m / ρ = 36 kg/h ÷ 1000 kg/m^3 = 0.036 m^3/h.
- Over 7 hours, the total melted volume is V = V̇ × t = 0.036 m^3/h × 7 h = 0.252 m^3.
- The pond is cylindrical with surface area A ≈ 1810 m^2 (as above). The thickness of the melted layer h is V / A ≈ 0.252 m^3 / 1810 m^2 ≈ 0.000139 m ≈ 1.39 × 10^−4 m ≈ 0.139 mm.
- To one significant figure, this thickness is about 0.1 mm.
- The candidate answer 0.06 mm is smaller than this calculation by about a factor of 2. If we were to force 0.06 mm, that would correspond to a total melted volume of A × h ≈ 1810 m^2 × 6.0 × 10^−5 m ≈ 0.1086 m^3, which over 7 hours would imply a melt rate of about 0.1086 m^3 / 7 h ≈ 0.0155 m^3/h, i.e., 15.5 L/h or about 15.5 kg/h, not the given 36 kg/h. The mismatch suggests either a miscalculation or a misprint in the 0.06 mm value.

Summary of evaluations:
- For part (a), using the given parameters yields a melt rate around 9 kg/hour, not 1.4×10^3 kg/hour. The 1.4×10^3 kg/hour option is inconsistent with the stated K, A, ΔT, and d.
- For part (b), the thickness is about 0.1–0.14 mm depending on rounding; the 0.06 mm figure is smaller than the calculation by roughly a factor of 2, indicating a possible error in that option.

Concluding observations: The physically derived results are: a) ~9 kg/hour (not 1400 kg/hour); b) ~0.1–0.14 mm (not 0.06 mm) for the melted layer after 7 hours. The provided numbers in the problem’s answer field appear inconsistent with the straightforward application of the given formulas and data.

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