Question

It takes 80 cal/g to change solid water at 0°C to liquid water. How many calories will it take to change 3 g of water at 0 °C to liquid water?

Answer 1

`"240 cal"`

Explanation:

The idea here is that the problem is providing you with a measure of how much heat is needed per gram of solid water, i.e. of ice, in order for a solid `->` liquid phase change to take place.

Simply put, you are given a measure of how much heat is required to melt `"1 g"` of ice at its normal melting point of `0^@"C"`.

So, you can say that `"80 cal g"^(-1)` is equivalent to saying that it takes `"80 cal"` of heat for every `"1 g"` of ice to convert it from solid at `0^@"C"` to lqiuid water at `0^@"C"`.

This means that `"3 g"` of ice would require

`3 color(red)(cancel(color(black)("g"))) * "80 cal"/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("240 cal")))`

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the mass of ice.

So remember this notation

`color(blue)("80 cal")color(red)("/g") " "=" " color(blue)("80 cal")color(white)(.)color(red)("g"^(-1))`

means that every `color(red)("1 g")` of ice at `0^@"C"` can be converted to liquid water at `0^@"C"` by providing it with `color(blue)("80 cal")` of heat.