Question

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? Why do we add 80 to the temperature difference?

Answer 1

From the 10.9 cm of ice,
Q = (10.9)(5000)(5000)(4.184)(1*4 + 80)
Q = 95,771,760,000 J ≈ 95.8 GJ

Assuming a linear temperature gradient from the bottom of the pond to the ice,

Q = (89.1)(5000)(5000)(4.184)(1*2)
Q = 18,639,720,000 J ≈ 18.6 GJ

so the total heat loss is approximated by
Q ≈ 114.4 GJ

This is the answer to the original question

and regarding your 2nd question about the necessity of adding 80 to the temperture difference

because here ice is converting into water so there is extra amount of energy that is called latent it is added to the heat of fusion of water is needed as the latent heat of fusion of ice is = 80 so it is added to the total heat