Question

How much ice at 0°C [Cp = 37.1 J/(mol ·°C), ΔH°fus = 6.01 kJ/mol] must be added to return the solution temperature to [ . . . ]?

The ΔH°soln of HNO3 is –33.3 kJ/mol. 11.0 mL of 12.0 M HNO3 is dissolved in 100.0 mL of distilled water initially at 25°C.

How much ice at 0°C [Cp = 37.1 J/(mol ·°C), ΔH°fus = 6.01 kJ/mol] must be added to return the solution temperature to 25°C after dissolution of the acid and equilibrium with the ice is reached? The molar heat capacity is 80.8 J/(mol·°C) for the solution, and the molar heat capacity is 75.3 J/(mol·°C) for pure water.

Answer 1

I got a mass of `"11.91 g"` of ice.

[Don't fall into the trap of using the same `barC_P` of water for the solution; the solution's specific heat capacity changed after mixing. You'd get `"12.63 g"` if you did that AND accidentally used the `barC_P` of ice for heating the ice AFTER it melted.

Speaking of which, do NOT use the heat capacity of ice... you'd get `"13.56 g"` if you did that but used the correct amount of heat transferred.]

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DISCLAIMER: REALLY LONG ANSWER!

First, we should determine the amount of heat transferred into the water, and thus its final temperature.

`11.0 cancel"mL" xx "1 L"/(1000 cancel"mL") xx "12.0 mol HNO"_3/cancel"L" = "0.132 mols HNO"_3`

The heat imparted is then:

`0.132 cancel"mols" xx "33.3 kJ"/cancel("mol HNO"_3) = "4.396 kJ" = "4396 J"`

This heat, assuming conservation of energy, goes into heating `"100.00 mL"` of water. If the water was added to the acid, then it would be heated by the acid dissolution using its own heat capacity.

Using a density of `"0.9970749 g/mL"` for water at `25^@ "C"`, and assuming a constant density (and step-function heat capacity, apparently) through the temperature change,

`q = n_"water"barC_(P,"water")DeltaT_"water"`

`"4396 J" = 100.00 cancel"mL" xx (0.9970749 cancel"g")/cancel"mL" xx cancel"1 mol"/(18.015 cancel"g") cdot "75.3 J/"cancel"mol"cdot""^@ "C" cdot (T_f - 25^@ "C")`

Solving this, the final temperature after the heating becomes:

`T_f = 35.55^@ "C"`

Now we want to return the solution to `25^@ "C"` by adding ice to it... We want the ice to melt and then heat up to `25^@ "C"`, because we are shooting for thermal equilibrium with the ice.

However, the solution has a higher heat capacity, of `"80.8 J/mol"cdot""^@ "C"`, and a new volume of `"111.00 mL"` (assuming additivity), with a new density of `"0.9938447 g/mL"`.

So, more heat than `"4396 J"` would be needed. The heat transfer follows conservation of energy as before:

`q_"soln" + q_"ice" = 0`

`overbrace(n_"soln"barC_(P,"soln")DeltaT_"soln")^"solution" + overbrace(q_1 + q_2)^"ice" = 0`

The heat that must be transferred out of the solution is:

`|q_"soln"| = 111.00 cancel"mL" xx (0.9938447 cancel"g")/cancel"mL" xx cancel"1 mol"/(18.015 cancel"g") cdot "80.8 J/"cancel"mol"cdotcancel(""^@ "C") cdot (25cancel(""^@ "C") - 35.55cancel(""^@ "C"))`

`= |-"5220 J"| = "5220 J"`

We want the total heat coming into the ice to be set up as follows:

`q_"ice" = q_1 + q_2`

`= overbrace(n_"ice"DeltabarH_(fus))^"melting" + overbrace(n_"ice"barC_(P,water)DeltaT_(water))^"heating melted ice"`

`= m_("ice")/"18.015 g/mol" cdot DeltabarH_(fus) + m_("ice")/"18.015 g/mol" cdot barC_(P,water)DeltaT_(water)`

`= m_("ice")/"18.015 g/mol"[DeltabarH_(fus) + barC_(P,water)DeltaT_(water)]`

We should be aware NOT to use `barC_P` of the ice... it is not ice after melting. From the conservation of energy equation,

`-q_"soln" = q_"ice"`

`=> "5220 J" = m_("ice")/"18.015 g/mol"[DeltabarH_(fus) + barC_(P,water)DeltaT_(water)]`

Evaluating the bracketed term first, we get:

`DeltabarH_(fus) + barC_(P,water)DeltaT_(water)`

`= (6.01 cancel"kJ")/"mol" xx ("1000 J")/cancel"1 kJ" + "75.3 J"//"mol"cdotcancel(""^@ "C") cdot (25cancel(""^@ "C") - 0cancel(""^@ "C"))`

`=` `"7892.5 J/mol"`

As a result, the mass of ice must be:

`color(blue)(m_("ice")) = (5220 cancel"J")/(7892.5 cancel"J"//cancel"mol") xx "18.015 g"/cancel"mol"`

`=` `color(blue)("11.91 g ice")`