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

##### 1 Answer

#### Explanation:

Your starting point here will be the **specific heat** of ice expressed in *joules per gram Kelvin*,

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

This tells you that in order to increase the temperature of **ice** by

Your goal here is to determine the specific heat of ice in *joules per kilogram Kelvin*,

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

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