A 100.0g ice cube at 0.0°C is placed in 650g of water at 25°C . what is the final temperature of the mixture?

1 Answer
Nov 26, 2015

Answer:

#11^@"C"#

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at #0^@"C"# to liquid at #0^@"C"#.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

#q_1 + q_2 = - q_3" " " "color(purple)((1))#, where

#q_1# - the heat absorbed by the solid at #0^@"C"#
#q_2# - the heat absorbed by the liquid at #0^@"C"#
#q_3# - the heat lost by the warmer water sample

The two equations that you will use are

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - heat absorbed/lost
#m# - the mass of the sample
#c# - the specific heat of water, equal to #4.18 "J"/("g" ""^@"C")#
#DeltaT# - the change in temperature, defined as final temperature minus initial temperature

and

#color(blue)(q = n * DeltaH_"fus")" "#, where

#q# - heat absorbed
#n# - the number of moles of water
#DeltaH_"fus"# - the molar heat of fusion of water, equal to #"6.01 kJ/mol"#

Use water's molar mass to find how many moles of water you have in the #"100.0-g"# sample

#100.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "5.551 moles H"_2"O"#

So, how much heat is needed to allow the sample to go from solid at #0^@"C"# to liquid at #0^@"C"#?

#q_1 = 5.551color(red)(cancel(color(black)("moles"))) * 6.01"kJ"/color(red)(cancel(color(black)("mole"))) = "33.36 kJ"#

This means that equation #color(purple)((1))# becomes

#"33.36 kJ" + q_2 = -q_3#

The minus sign for #q_3# is used because heat lost carries a negative sign.

So, if #T_"f"# is the final temperature of the water, you can say that

#"33.36 kJ" + m_"sample" * c * DeltaT_"sample" = - m_"water" * c * DeltaT_"water"#

More specifically, you have

#"33.36 kJ" + 100.0color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 0)color(red)(cancel(color(black)(""^@"C"))) = -650color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 25)color(red)(cancel(color(black)(""^@"C")))#

#"33.36 kJ" + "418 J" * (T_"f" - 0) = - "2717 J" * (T_"f" - 25)#

Convert the joules to kilojoules to get

#33.36 color(red)(cancel(color(black)("kJ"))) + 0.418 color(red)(cancel(color(black)("kJ"))) * T_"f" = -2.717 color(red)(cancel(color(black)("kJ"))) * (T_"f" - 25)#

This is equivalent to

#0.418 * T_"f" + 2.717 * T_"f" = 67.925 - 33.36#

#T_"f" = 34.565/(0.418 + 2.717) = 11.026^@"C"#

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

#T_"f" = color(green)(11^@"C")#