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`

Step-by-step 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* volume =5.50* 5.0=27.5g`

Volume of water = 25.0 ml

Density of metal = 1.0 g/ml

Mass of metal =
`density* volume =1.0* 25.0=25.0g`

`heat_(absorbed)=heat_(released)`

As we know that,

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`-[m_1* c_1* (T_(final)-T_1)]=[m_2* c_2* (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* c_1* (41.0-153)^0C=[25.0g* 4.814J/g^0C* (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`