Question

Specific heat of ice in `"J/kg K"`?

Answer 1

`"2060 J kg"^(-1)"K"^(-1)`

Explanation:

Your starting point here will be the specific heat of ice expressed in joules per gram Kelvin, `"J g"^(-1)"K"^(-1)`, which is listed as being equal to

`c_"ice" = "2.06 J g"^(-1)"K"^(-1)`

This tells you that in order to increase the temperature of `"1 g"` of ice by `"1 K"` you must provide it with `"2.06 J"` of heat.

Your goal here is to determine the specific heat of ice in joules per kilogram Kelvin, `"J kg"^(-1)"K"^(-1)`, which essentially tells you how much heat is required in order to increase the temperature of `"1 kg"` of ice by `"1 K"`.

The conversion factor that takes you from grams to kilograms is

`color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))`

You can use this conversion factor to get

`2.06"J"/(color(red)(cancel(color(black)("g"))) * "K") * (10^3color(red)(cancel(color(black)("g"))))/("1 kg") = color(green)(|bar(ul(color(white)(a/a)color(black)(2.06 * 10^3color(white)(a) "J kg"^(-1)"K"^(-1))color(white)(a/a)|)))`

So, in order to increase the temperature of `"1 kg"` of ice by `"1 K"`, you must provide it with

`2.06 * 10^3"J" = "2060 J"`