Question

an irregular lump of an unknown metal has a measured density of 5.46 g/ml. the metal is heated to a temperature of 187°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 34.9 ml, and the temperature is recorded as 39.7°c. what is the specific heat of the unknown metal sample? assume no heat is lost to the surroundings.

Explanation:

Step1: Calculate the volume and mass of the metal

The volume of the metal \(V = 34.9\space mL - 25.0\space mL=9.9\space mL\)
Using the density formula \(m=
ho V\), with \(
ho = 5.46\space g/mL\) and \(V = 9.9\space mL\), we get \(m=5.46\times9.9 = 54.054\space g\)

Step2: Calculate the heat absorbed by water

The specific heat capacity of water \(c_{water}=4.184\space J/(g\cdot^{\circ}C)\), the mass of water \(m_{water}=
ho_{water}V_{water}\), with \(
ho_{water} = 1\space g/mL\) and \(V_{water}=25.0\space mL\), so \(m_{water}=25.0\space g\)
The temperature change of water \(\Delta T_{water}=39.7 - 25.0=14.7^{\circ}C\)
Using the heat formula \(Q = mc\Delta T\), the heat absorbed by water \(Q_{water}=25.0\times4.184\times14.7 = 1530.18\space J\)

Step3: Calculate the temperature change of the metal

The temperature change of the metal \(\Delta T_{metal}=187 - 39.7 = 147.3^{\circ}C\)

Step4: Calculate the specific heat of the metal

Since \(Q_{metal}=-Q_{water}\) (heat lost by metal = heat gained by water), and \(Q = mc\Delta T\)
\(c_{metal}=\frac{Q_{metal}}{m_{metal}\Delta T_{metal}}=\frac{- 1530.18}{54.054\times147.3}\)
\(c_{metal}=0.193\space J/(g\cdot^{\circ}C)\)

Answer:

\(0.193\space J/(g\cdot^{\circ}C)\)