Answer:
Heat required to melt 1 lb of ice is 151.469 KJ
Explanation:
We have given mass of ice = 1 lb
We know that 1 lb = 0.4535 kg
Latent heat of fusion for ice =334 KJ/kg
Amount if heat for fusion of ice is given by
[tex]Q=mL[/tex], here m is mass of ice and L is latent heat of fusion
So heat [tex]Q=mL=0.4535\times 334=151.469kj[/tex]
So heat required to melt 1 lb of ice is equal to 151.469 KJ