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?

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

see earlier answer

massxgravity x heght=weightxheight= pe

4900=490xh

h=10