What is the specific heat of the unknown metal sample?

An irregular lump of an unknown metal has a measured density of 2.97 g/mL. The metal is heated to a temperature of 173 °C and placed in a graduated cylinder filled with 25.0 mL of water at 25.0 °C. After the system has reached thermal equilibrium, the volume in the cylinder is read at 34.0 mL, and the temperature is recorded as 40.6 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

2 Answers

#c=0.46 J/(g*°C)#

Explanation:

Write down given information for both substances (mass, specific heat, change in temperature)

For Water :

#m = 25.0 g#

#c = 4.181 J/(g*°C)#

#T_(i nitial) = 25°C#

#T_(fi nal) = 40.6°C#

#DeltaT = T_(fi nal)-T_(i nitial) = 40.6°C - 25°C = 15.6°C#

For Metal :

#m = D*V = 2.97 g/mL * 9mL = 26.73 g#

#c = ?#

#T_(i nitial) = 173°C#

#T_(fi nal) = 40.6°C#

#DeltaT = T_(fi nal)-T_(i nitial) = 40.6°C - 173°C = - 132.4°C#

Use the formula #Q=mcDeltaT# to find the change in energy in water.

#Q_w=(25.0 g)(4.181 J/(g*°C))(15.6°C)#
#Q_w=1630.59 J#

This (#Q_w#) is the amount of energy the water gained, so this means the metal lost the same amount(#-Q_m#), according to the law of conservation of energy.

#Q_w=-Q_m#

Rearrange the formula #Q=mcDeltaT# to find #c# of the metal.

#c=Q_m/(m*DeltaT)#

#c=(-1630.59J)/((26.73 g)(- 132.4°C))#

#c=0.46 J/(g*°C)#

Feb 4, 2018

#C_m ~~ "0.457 J/g"^@ "C"#


The displacement of water gives the volume and thus mass of the metal:

#"34.0 - 25.0 mL" xx "2.97 g"/"mL" = "26.73 g"#

By conservation of energy,

#q_m + q_w = 0#

where #m# is metal and #w# is water.

Thus

#q_m = -q_w#

and

#m_mC_mDeltaT_m = -m_wC_wDeltaT_w#

where #C# is the constant-pressure specific heat capacity in #"J/g"^@ "C"#, #m_i# the mass, and #DeltaT# the change in temperature in #""^@ "C"#.

They both reach the same thermal equilibrium temperature, so

#m_mC_m(T_f - T_m) = -m_wC_w(T_f - T_w)#

Therefore, assuming the specific heat capacity of water stays constant in the temperature range, and that its density is that at #40.6^@ "C"#:

#color(blue)(C_m) = -(m_wC_w(T_f - T_w))/(m_m(T_f - T_m))#

#= -("25.0 mL" xx "0.9919880 g"/"mL" cdot "4.184 J/g"^@ "C" cdot (40.6^@ "C" - 25.0^@ "C"))/("26.73 g" cdot (40.6^@ "C" - 173^@ "C"))#

#= color(blue)("0.457 J/g"^@ "C")#