Question

How much heat is required to convert 6.83 g of ice at − 12.0 ∘C to water at 20.0 ∘C ? (The heat capacity of ice is 2.09 J/g∘C , ΔHvap(H2O)=40.7kJ/mol , ΔHfus(H2O)=6.02kJ/mol ) Express your answer with the appropriate units.

Answer 1

Step-by-step explanation

Given that:

`\begin{gathered} \text{ The mass of ice }=\text{ 6.83g} \\ \text{ initial temperature }=-12\degree C \\ \text{ Final temperature }=20\degree C \\ \text{ Specific heat capacity of ice }=\text{ 2.09 J/g}\degree C \\ \text{ }\Delta H_(Vapor)(H_2O)\text{ }=\text{ 40.7kJ/mol} \\ \text{ }\Delta H_(Fusion)(H_2O)\text{ }=\text{ 6.02 kJ/mol} \end{gathered}`

To find the heat required to convert ice to water, follow the steps below

Step 1: Write the heat formula

`q\text{ }=\text{ mc}\Delta\theta`

Since, the ice will be converted to solid first, the the final final temperature will be 0 degrees Celcius

`\begin{gathered} \text{ q }=\text{ 6.83}*2.09*\text{ \lbrack\lparen0-\lparen-12\rparen\rbrack} \\ \text{ q }=\text{ 6.83}*2.09*\text{ \lparen0 }+12) \\ \text{ q}=\text{ 6.83}*\text{ 2.09}*\text{ 12} \\ \text{ q }=\text{ 171.2964 J} \end{gathered}`

Step 2: find the heat required to convert the solid to liquid from 0 degrees Celcius to 0 degrees Celcius

`\text{ q}=\text{ m }*\text{ }\Delta H_(fus)`

Where m is the mass of the sample