A local ice hockey pond is at 4°C when the air temperature falls, causing the water temperature to drop to 0°C with 10.9 cm of ice forming at the surface at a temperature of 0°C. How much heat was lost if the unfrozen pond is 1 m deep and 50 m on each side?

Question

Physics

Kyan Holmes

22 April, 02:33

A local ice hockey pond is at 4°C when the air temperature falls, causing the water temperature to drop to 0°C with 10.9 cm of ice forming at the surface at a temperature of 0°C. How much heat was lost if the unfrozen pond is 1 m deep and 50 m on each side?

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Answers (1)

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Kasen Jarvis · Answer 1

Answer: 114.4 GJ

Explanation:

Heat loss Q=U*A*ΔT

Heat loss of size A is determined by the U value of materials and the difference in temperature.

From 10.9cm from the ice

50m = 5000cm

A = 5000*5000

Q= = (10.9) (5000) (5000) (4.184) (1*4 + 80)

Q = 95,771,760,000J

Q≈ 95.8 GJ

Linear gradient from the bottom of the pond to the ice:

Q = (89.1) (5000) (5000) (4.184) (1*2)

Q = 18,639,720,000J

Q ≈ 18.6 GJ

Total heat loss:

Q = 95.8GJ + 18.6GJ

Q = 114.4 GJ