Question

A sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C^−1 , is put into a calorimeter (see sketch at right) that contains 300.0g of water. The aluminum sample starts off at 94.5°C and the temperature of the water starts off at 21.0°C .When the temperature of the water stops changing it's 23.8°C .The pressure remains constant at 1atm .Calculate the mass of the aluminum sample.

Answer 1

Answer: The mass of aluminium sample is 55.4 gram

Explanation:

$Q_{absorbed}=Q_{released}$

As we know that,  

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

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$      

where,

$m_1$ = mass of aluminium = ?

$m_2$ = mass of water = 300.0 g

$T_{final}$ = final temperature = $23.8^0C$

$T_1$ = temperature of aluminium = $94.5^oC$

$T_2$ = temperature of water = $21.0^oC$

$c_1$ = specific heat of aluminium = $0.897J/g^0C$

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

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

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

$-[m_1\times 0.897\times (23.8-94.5)^0C]=[300.0g\times 4.184\times (23.8-21.0)]$

$m_1=55.4g$

Therefore, the mass of aluminium sample is 55.4 gram