Question

What is the final temperature of the combined system when `"10.0 g"` of ice is placed into `"50.0 g"` of water at `32.0^@ "C"`? `c_"water" = "4.184 J/g"^@ "C"`, while `DeltaH_(fus)^@ = "333 J/g"` for ice melting into water.

Answer 1

`"Final temperature" ~~ 13.39^oC`

Explanation:

According to principle of calorimetry,

`"Heat absorbed by ice" + "Heat absorbed by water formed from ice" = "Heat lost by hot water"`

`10xx333 + 10xx4.18xx(T_f-0) = 50xx4.18xx(32-T_f)`

solving for `T_f` ,

`T_f ~~ 13.39^oC`

Answer 2

I got `13.40^@ "C"`.

---

Since ice is put into hotter water, it must first melt. The total heat contribution to melt the ice is:

`color(green)(q_"ice" = m_"ice"DeltabarH_"fus")`

where `DeltabarH_"fus"` is the mass enthalpy of a phase transition in `"J/g"`, and `m` is the grams of water.

...and its value is:

`q_"melt" = 10.0 cancel"g ice" xx "333 J"/cancel"g ice" = "3330 J"`

The contribution to heat it up to some final temperature is going to occur at constant atmospheric pressure but nonconstant temperature, so

`color(green)(q_"ice" = overbrace(m_"ice")^("as melted water")c_"water"overbrace(DeltaT_"ice")^"as melted water")`

We are saying:

`overbrace("ice")^(0.00^@ "C") stackrel("Based on "DeltaH_"fus"" ")(->) overbrace("water")^(0.00^@ "C") stackrel("Based on specific heat capacity"" ")(->) overbrace("water")^(T_f)`

And the contribution to cooling the hotter water has a magnitude of:

`color(green)(q_"water" = m_"water"c_"water"DeltaT_"water")`

By conservation of energy, the thermal energy from the hotter water goes into melting the ice and into heating it. So:

`q_"melt" + q_"ice" + q_"water" = 0`

`=> q_"melt" + q_"ice" = -q_"water"`

`=> m_"ice"DeltabarH_"fus" + overbrace(m_"ice")^("as melted water")c_"water"overbrace(DeltaT_"ice")^"as melted water"`

`= -m_"water"c_"water"DeltaT_"water"`

Expanding this out, we then get:

`"3330 J" + "10.0 g ice" cdot "4.184 J/g"^@ "C" cdot (T_f - 0.00^@ "C")`

`= -"50.0 g water" cdot "4.184 J/g"^@ "C" cdot (T_f - 32.0^@ "C")`

We know the units will work out since this is all in `"J"`, `"g"`, and `""^@ "C"`, and it turns out that distributing terms is easier than working this out algebraically with no numbers.

We now omit the units on purpose and we'll put them back later.

`3330 + 41.84T_f = -209.2(T_f - 32)`

Distribute terms to get:

`3330 + 41.84T_f + 209.2T_f = 6694.4`

Solve for `T_f`, the equilibrium temperature (which MUST be between `0.00^@ "C"` and `32.00^@ "C"`; why?):

`3330 + 251.04T_f = 6694.4`

`=> 251.04T_f = 3364.4`

`=> color(blue)(T_f) = (3364.4 cancel"J/g")/(251.04 cancel"J/g"cdot""^@ "C")`

`= ulcolor(blue)(13.40^@ "C")`