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

Jun 17, 2016

${\text{2060 J kg"^(-1)"K}}^{- 1}$

#### Explanation:

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

${c}_{\text{ice" = "2.06 J g"^(-1)"K}}^{- 1}$

This tells you that in order to increase the temperature of $\text{1 g}$ of ice by $\text{1 K}$ you must provide it with $\text{2.06 J}$ of heat.

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

The conversion factor that takes you from grams to kilograms is

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 kg" = 10^3"g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 $\text{1 kg}$ of ice by $\text{1 K}$, you must provide it with

$2.06 \cdot {10}^{3} \text{J" = "2060 J}$