Question

Question #6033f

Answer 1

Here's how you can do that.

Explanation:

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"))`