Question

To a sample of water at 23.4oC in a constant pressure calorimeter of negligible heat capacity is added a 12.1 g piece of aluminium whose temperature is 81.7oC. If the final temperature of water is 24.9 oC, calculate the mass of the water in the calorimeter. Ans:98.6g

-I know that The specific heat of aluminum is 0.900 J/g ‡ ÁC
- _T Al is 24.9ÁC _ 81.7ÁC = _56.8ÁC
- _Twater and _Tcalorimeter are both 24.9ÁC _ 23.4ÁC = 1.5ÁC.
-The specific heat of water is 4.184 J/g ‡ ÁC.
But I tried using m=q/s_t. I'm really stuck,can anyone help me?

Answer 1

We know that the thermal energy lost by the hot piece of aluminum will be gained by the water. Another thing we know is that the final temperature of water is achieved at thermal equilibrium, this means that this final temperature is also that of the piece of aluminum within the water.
Energy lost by aluminum:
q = mcΔt
= 12.1 x 0.9 x (81.7 - 24.9)
= 618.552 Joules
Equating this to the energy gained by water:
618.552 = m x 4.184 x (24.9 - 23.4)
m = 98.553
=98.6 grams

Answer 2

Answer: 98.6 grams

Explanation:

$heat_{released}=heat_{absorbed}$

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 aluminium = 12.1 g

$m_2$ = mass of water = ?

$T_{final}$ = final temperature = $24.9^0C$

$T_1$ = temperature of aluminium = $81.7^oC$

$T_2$ = temperature of water = $23.4^oC$

$c_1$ = specific heat of aluminium = $0.900J/g^0C$

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

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

$-12.1\times 0.900\times (24.9-81.7)=-[m_2\times 4.184\times (24.9-23.4)]$

$m_2=98.6g$

Therefore, the mass of the water in the calorimeter is 98.6 grams