# How much heat is required to convert 5.88 g of ice at -12.0°C to water at 25.0°C, if the heat capacity of ice is 2.09 J/°C?

Jan 2, 2016

${q}_{\text{total" = "2.73 kJ}}$

#### Explanation:

Your strategy here will be to determine

• how much heat is required to convert that sample of ice from solid at $- {12.0}^{\circ} \text{C}$ to solid at ${0}^{\circ} \text{C}$

• determine how much heat is required to convert that sample of ice from solid at ${0}^{\circ} \text{C}$ to liquid at ${0}^{\circ} \text{C}$

• finally, determine how much heat is required to heat that sample of now liquid water from ${0}^{\circ} \text{C}$ to ${25.0}^{\circ} \text{C}$

In order to be able to calculate all this, you will need to know the values for the specific heat of ice and that of water, and the value of the enthalpy of fusion, $\Delta {H}_{\text{fus}}$, for water

${c}_{\text{ice" = 2.09"J"/("g" ""^@"C")" }}$ and " "c_"water" = 4.18"J"/("g" ""^@"C")

$\Delta {H}_{\text{fus" = 334"J"/"g}}$

http://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html

Two equations will come in handy here

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - heat absorbed / lost
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

color(blue)(q = m * DeltaH_"fus")" ", where

$q$ - heat absorbed / lost
$m$ - the mass of the sample
$\Delta {H}_{\text{fus}}$ - the enthalpy of fusion of the substance

So, to get your $\text{5.88-g}$ sample of ice from solid at $- {12.0}^{\circ} \text{C}$ to ${9}^{\circ} \text{C}$, you will need to provide it with

${q}_{1} = 5.88 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 2.09"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * [0 - (-12.0)]color(red)(cancel(color(black)(""^@"C}}}}$

${q}_{1} = \text{147.47 J}$

To get you sample to undergo a solid $\to$ liquid phase change at its melting point, i.e. at ${0}^{\circ} \text{C}$, you will need to provide it with

${q}_{2} = 5.88 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 334"J"/color(red)(cancel(color(black)("g}}}}$

${q}_{2} = \text{1963.9 J}$

Finally, to heat the sample from ${0}^{\circ} \text{C}$ to ${25.0}^{\circ} \text{C}$, you need to provide it with

${q}_{3} = 5.88 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (25.0 - 0)color(red)(cancel(color(black)(""^@"C}}}}$

${q}_{3} = \text{614.46 J}$

The total heat* required to get your sample from ice at $- {12.0}^{\circ} \text{C}$ to liquid at ${25.0}^{\circ} \text{C}$ will thus be equal to

${q}_{\text{total}} = {q}_{1} + {q}_{2} + {q}_{3}$

${q}_{\text{total" = "147.47 J" + "1963.9 J" + "614.46 J}}$

${q}_{\text{total" = "2725.83 J}}$

Rounded to three sig figs and expressed in kilojoules, the answer will be

q_"total" = color(green)("2.73 kJ")