# 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

#### Answer:

#### 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 *liquid* at

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

The two equations that you will use are

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

*final temperature* minus *initial temperature*

and

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

*number of moles* of water

*molar heat of fusion* of water, equal to

Use water's molar mass to find how many *moles* of water you have in the

#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 *liquid* at

#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

#"33.36 kJ" + q_2 = -q_3#

The minus sign for **heat lost** carries a negative sign.

So, if

#"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")#