# 10.0g of a metal, initially at 25 Celsius, are placed into 10.0 g of water, initially at 100 Celsius. Which metal will have the lowest final temperature? A) Aluminium (0.902 J/g*Celsius) B) Gold (0.129 J/g* Celsius) C) Copper (0.385 J/g*Cesius)

Aluminum will produce the lowest final temperature at ##”86.7″^(“o”)”C”##.

##Q=mc####DeltaT##, where ##Q## is the amount of heat energy lost or gained, ##m## is the mass in grams, and ##DeltaT## is the change in temperature, ##T_”final” – T_”initial”##.

The final temperature occurs when the metal and water temperatures are the same, meaning they are at equilibrium. At equilibrium, the magnitude of the heat energy gained by the metal is equal to the magnitude of the heat energy lost by the water. The of water is ##”4.184 J/g”^(“o”)”C”##.

In order to find the final temperature, or the temperature at equilibrium, set the heat capacity for the metal and the water equal, except Q for water is negative because it loses heat energy.

##Q_”metal” = -Q_”water”## (The negative sign indicates that the water loses heat energy.)

##mc(T_f – T_i) = -mc(T_f – T_i)##

A) Aluminum

##10.0g*0.902 “J/g^oC”*(T_f – 25^oC) = -10.0g*4.184 “J/g*”^oC”*(T_f – 100^oC)##

##9.02(T_f – 25^oC) = -41.84(T_f – 100^oC)##

Apply the distributive property.

##9.02T_f – 225.5^oC = -41.84T_f + 4184^oC##

##50.86T_f = 4410^oC##

Divide both sides by ##50.86##.

##T_f = 86.7^oC##

B) Gold

##10.0g*0.129 “J/g^oC”*(T_f – 25^oC) = -10.0g*4.184 “J/g*”^oC”*(T_f – 100^oC)##

##1.29(T_f – 25^oC) = -41.84(T_f – 100^oC)##

Apply the distributive property.

##9.02T_f – 32.25^oC = -41.84T_f + 4184^oC##

##43.13T_f = 4216^oC##

Divide both sides by ##43.13##.

##T_f = 97.8^oC##

C) Copper

##10.0g*0.385 “J/g^oC”*(T_f – 25^oC) = -10.0g*4.184 “J/g*”^oC”*(T_f – 100^oC)##

##3.85(T_f – 25^oC) = -41.84(T_f – 100^oC)##

Apply the distributive property.

##3.85T_f – 96.25^oC = -41.84T_f + 4184^oC##

##45.69T_f = 4280^oC##

Divide both sides by ##45.69##.

##T_f = 93.7^oC##

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