Question

An irregular lump of an unknown metal has a measured density of 5.50 g/mL. The metal is heated to a temperature of 153 °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 30.0 mL, and the temperature is recorded as 41.0 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

Answer 1

Answer: $0.543J/g^0C$

Explanation:

Volume of metal = volume of water displaced = (30.0 - 25.0) ml = 5.0 ml

Density of metal = 5.50 g/ml

Mass of metal =$density\times volume =5.50\times 5.0=27.5g$

Volume of water = 25.0 ml

Density of metal = 1.0 g/ml

Mass of metal =$density\times volume =1.0\times 25.0=25.0g$

$heat_{absorbed}=heat_{released}$

As we know that,  

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)]$         .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 27.5 g

$m_2$ = mass of water = 25.0 g

$T_{final}$ = final temperature = ?$41.0^0C$

$T_1$ = temperature of metal = $153^oC$

$T_2$ = temperature of water = $25.0^oC$

$c_1$ = specific heat of lead = ?

$c_2$ = specific heat of water= $4.184J/g^0C$

Now put all the given values in equation (1), we get

$27.5g\times c_1\times (41.0-153)^0C=[25.0g\times 4.814J/g^0C\times (41.0-25.0)^0C]$

$c_1=0.543J/g^0C$

Thus the specific heat of the unknown metal sample is $0.543J/g^0C$