Question

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of water at a temperature of 23.0°C. After stirring, the final temperature of both copper and water is 26.0°C. Assuming no heat losses, and that the specific heat capacity of water is 4.184 J/(g·°C), what is the molar heat capacity of aluminum, Cm(Al)?

Answer 1

Answer: The molar heat capacity of aluminum is
`25.3J/mol^0C`

Step-by-step explanation:

`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 water = 130.0 g

`m_2` = mass of aluminiunm = 23.5 g

`T_(final)` = final temperature =
`26.0^oC=(273+26)K=299K`

`T_1` = temperature of water =
`23^oC=(273+23)K=296K`

`T_2` = temperature of aluminium =
`100^oC=273+100=373K`

`c_1` = specific heat of water=
`4.184J/g^0C`

`c_2` = specific heat of aluminium= ?

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

`130.0* 4.184* (299-296)=-[23.5* c_2* (299-373)]`

`c_2=0.938J/g^0C`

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity =
`0.938J/g^0C* 27g/mol=25.3J/mol^0C`