A 25-g piece of an unknown metal is heated to 100°c and then placed into a calorimeter containing 45g of water at a temperature of 23.0°c.The final temperature of both is 27.0°c. What is the specific heat of the unknown metal? Show the steps

1 Answer
Rita
May 13, 2018

#-0.41J/(g°C)# = #c#

Explanation:

First we use the equation #q# = #mcDeltaT# to find the amount of heat (measured in Joules) that the water gained

#m# (mass) = #45g#
#c# (specific heat for water) = #(4.2J)/(g°C)#
#DeltaT# (temp. change) = #(27.0°C - 23°C)#

  • #DeltaT# = (Final temp. - Initial temp.)

#q# = #(45g)((4.2J)/(g°C))(27.0°C - 23°C)#

  • subtract the temperatures then multiply everything

#q# = #(45cancel(g))((4.2J)/(cancel(g°C)))(4cancel(°C))#

  • The #g# and #°C# units cancel, leaving #J#

#q# = #756J#

Now we take the same equation #q# = #mcDeltaT# and plug in values for solving the specific heat for the metal

#q# = #756J#
#m# = #25g#
#c# = unknown
#DeltaT# = #(27.0°C - 100°C)#

#756J# = #(25g)(c)(27.0°C - 100°C)#

  • subtract the temps, then divide each side by #(25g)(-73°c)#

#(756J)/((25g)(-73°c))# = #((cancel(25g))(c)(cancel(-73°C)))/((cancel(25g))(cancel(-73°c)))#

#(756J)/((25g)(-73°c))# = #c#

  • Multiply #25g# by #-73°c#

#(756J)/(-1825g°C)# = #c#

  • Then divide #756J# by #-1825g°C# to get the answer in #J/(g°C)#

#-0.41J/(g°C)# = #c#

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