# 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.

##### 2 Answers

#### Explanation:

According to principle of calorimetry,

solving for

I got

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

**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 **equilibrium temperature** (which MUST be between

#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")#