A 134 g stainless steel ball bearing (sp ht = 0.50 J g-1 °C-1) at 525°C is dropped onto a large block of ice a

Question

A 134 g stainless steel ball bearing (sp ht = 0.50 J g-1 °C-1) at 525°C is dropped onto a large block of ice at 0°C, what mass of liquid water will form? (H°fusion = 6.01 kJ/mol H2O(s))

Answer 1

Heat released by the ball bearing when its temperature drops from 525°C to 0°C is:
134*0.50*525 = 35.2 kJ.
35.2 kJ / (6.01 kJ/mol) is the number of moles of ice would change to water. Remember the molar mass of water is 18g/mol. So the final result is:
18.0*35.2/6.01 = 105g = 0.11kg.