Question #6033f

For starters, you know that in order to melt #"100 g"# of ice at its normal melting point, i.e. to go from ice at #0^@"C"# to liquid water at #0^@"C"#, you need

#100 color(red)(cancel(color(black)("g"))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = "33,355 J"#

This is the case because the enthalpy of fusion of water, #DeltaH_"fus"#, which tells you the energy needed to convert #"1 g"# of ice at #0^@"C"# to liquid water at #0^@"C"#, is equal to #"333.55 J g"^(-1)#.

This tells you that in order to convert #"1 g"# of ice at #0^@"C"# to liquid awter at #0^@"C"#, you need to provide #"333.55 J"# of heat.

Now, you only have #"2,000 J"# of energy available, so right from the start, you can say for a fact that not all the ice will be converted to liquid water at #0^@"C"#.

#overbrace("33,355 J")^(color(blue)("what you need")) " " > " " overbrace("2,000 J")^(color(blue)("what you have"))#

To find the mass of ice that would melt, use the enthalpy of fusion again.

#2000 color(red)(cancel(color(black)("J"))) * "1 g"/(333.55color(red)(cancel(color(black)("J")))) = "5.996 g"#

This means that only

#"5.996 g " ~~ " 6 g"#

of ice will melt, the rest will remain as ice, not as liquid water. The final temperature of the ice + liquid water mixture will be #0^@"C"# because you didn't provide enough heat to

1. Melt all the ice to get liquid water at #0^@"C"#
2. Warm the liquid water to a temperature that is higher than #0^@"C"#

So you only converted #"6 g"# of ice at #0^@"C"# to #"6 g"# of liquid water at #0^@"C"#, i.e. you performed a solid #-># liquid phase change, while leaving #"94 g"# of ice at #0^@"C"#.

#overbrace("100 g ice at 0"^@"C")^(color(blue)("ice at 0"^@"C"))" " stackrel(color(white)(acolor(red)("+ 2,000 J")aaa))(->) overbrace(" 6 g ")^(color(blue)("liquid water at 0"^@"C")) + overbrace(" 94 g ")^(color(blue)("ice at 0"^@"C"))#