91Ó°ÊÓ

Heat given off by copper = - heat absorbed by water

Let’s assume heat = q

q_copper = - q_water

As we know that heat is related to mass, specific heat, and temperature which is given as follows.

\({{\rm{(c \times m \times \Delta T)}}_{{\rm{metal}}}}{\rm{\; = \;\;(c \times m \times \Delta T}}{{\rm{)}}_{{\rm{Water}}}}\)

Let f = final and i = initial terms and by expanding the above equation, the new equation becomes:

\({{\rm{C}}_{{\rm{metal}}}}{\rm{ \times }}{{\rm{m}}_{{\rm{metal}}}}{\rm{ \times (Tf}}{{\rm{,}}_{{\rm{metal}}}}{\rm{--Ti}}{{\rm{,}}_{{\rm{metal}}}}{{\rm{)}}_{}}{\rm{\; = \;\;}}{{\rm{C}}_{{\rm{water}}}}{\rm{ \times }}{{\rm{m}}_{{\rm{water}}}}{\rm{ \times (Tf}}{{\rm{,}}_{{\rm{water}}}}{\rm{--Ti}}{{\rm{,}}_{{\rm{water}}}}{\rm{)}}\)

From the data given

• Initial temperature of metal = 178 ËšC.
• Initial temperature of water = 24 ËšC.
• Final temperature of metal and water = 29.7 ËšC.
• Mass of metal = 92.9 g
• 75 mL water = 75 g

The knowns values are substituted.

\(\begin{aligned}{}{C_{metal}}\; \times {\rm{ }}92.9{\rm{ }}g{\rm{ }} \times {(29.7^\circ C--178^\circ C)_{}}\; &= \; - \;4.18{\rm{ }} \times {\rm{ }}75{\rm{ }}g{\rm{ }} \times {\rm{ }}(24^\circ C--29.7^\circ C)\\{C_{metal}} &= \frac{{1786.95}}{{13777.07}}\\{C_{metal}}_{\;\;}\; &= {\rm{ }}0.1297 \sim \;0.13J/g^\circ C\end{aligned}\)

\({C_{metal}}_{\;\;}\; = 0.13J/g^\circ C\) and this specific heat is close to either gold or lead.

But solely by numerical values, we can’t determine which metal it is. However, the observation that the silver/grey metal in addition to the values of specific heat indicates the metal is lead.