To get this answer, we'll need to split the process into #3# parts:
- Ice getting to #0°C# from #-50.0 °C#.
- Ice becoming water at #0°C#.
- Water getting to #70°C# from #0°C#.
1: For the first step, we need to use #q=mCDeltaT# to calculate the enthalpy change.
We know that the specific heat of ice, #C#, is #"2.06 J/g°C"#. We also know the temperature change—#50.0°C#.
We don't know the mass, but we can find it because the question told us that we have #1.00# mole of ice. The chemical formula for ice, like water, is #H_2O#, therefore the mass of #1# mole is:
#2*"1 mole of H" + "1 mole of O" = 2*"1.008g" + "16.00g" = "18.02g"#
Now that we know the mass, we can just plug it into the formula and solve for #q#, or enthalpy change for the first step.
#q=mCDeltaT = 18.02 * 2.06 * 50.0 = 1860J#
2: For the second step, we need to find the amount of energy it takes for #"18.02g"# of ice to melt. To do this, we just multiply the mass by the heat of fusion (basically, the energy it takes to melt one gram of something).
The heat of fusion for water is #"333J/g"#
#DeltaH = 18.02 * 333 = 6001J#
3: Lastly, we can apply #q=mCDeltaT# again. Our mass is #"18.02g"#, specific heat capacity of water is #"4.184J/g°C"#, and the temperature change is #70.0°C#.
#q = mCDeltaT = 18.02 * 4.184 * 70 = 5278J#
After adding all of this up, we get the total enthalpy change:
#1860 + 6001 + 5278 = 13100J#
