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
#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")#