heat ice to 0ºC ... 10.0 g *10.0 ºC * 2.09 J/gºC
melt ice ... 10.0 g * 334 J/g
heat water to 100ºC ... 10.0 g * 4.184 J/gºC * 100.0 ºC
boil water ... 10.0 g * 2.23 kJ/g
heat steam to 115ºC ... 10.0 g * 1.84 J/gºC * 15.0 ºC
add the steps
How much heat in kilojoules is required to warm 10.0 g of ice, initially at -10.0 ∘C, to steam at 115 ∘C. The heat capacity of ice is 2.09 J/g∘C and that of steam is 1.84 J/g∘C.
2 answers
2.83 kJ