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")