Answer:
the final temperature of the system of the system is 25.32°C
Explanation:
We are not given specific capacity of water and aluminium, so we use their standard values, also we are not given the density if water so we assume the standard vale of density of water
The aluminium calorimeter has a mass Mc= 120g
Volume of water in calorimeter = 100ml at θc =20°C
Density of water is
1000Kg/m³ = 1g/mL
Then, density = mass/ volume
Mass=density ×volume
Mass=1g/mL×100mL
Mass=100gram
Then, the mass of water is
Mw = 100gram
Mass of brass is Mb = 100gram
The temperature of brass is θb=100°C
The specific heat capacity of water is Cw= 1cal/g°C
The specific heat capacity of aluminum Ca=0.22cal/g°C
We are looking for final temperature θf=?
Given that the specific heat capacity of brass is Cb=0.09Cal/g°C
Using the principle of calorimeter;
The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
So, the calorimeter gained heat and the liquid in the calorimeter gain heat too
Heat gain by calorimeter(Hc) = Mc•Ca•∆θ
Where Mc is mass of calorimeter,
Ca is Specific Heat capacity of Calorimeter
∆θ=(θf-θc)
Hc=Mc•Ca•∆θ
Hc=120•0.22•(θf-20)
Hc=26.4(θf-20)
Hc=26.4θf-528
Also, heat gain by the water
Heat gain by wayer(Hw) = Mw•Cw•∆θ
Where Mw is mass of water,
Cw is Specific Heat capacity of water
∆θ=(θf-θw),
Note that the temperature of the water and the calorimeter are the same at the beginning i.e. θc=θw=20°C
Hw=Mw•Cw•∆θ
Hw=100•1•(θf-20)
Hw=100(θf-20)
Hw=100θf-2000
Also heat loss by the brass is given by
heat loss by brass
Heat loss by brass(Hb)= Mb•Cb•∆θ
Where Mb is mass of brass,
Cb is Specific Heat capacity of brass
∆θ=(θb-θf)
Therefore,
Hb=Mb•Cb•∆θ
Hb=100•0.09•(100-θf)
Hb=9(100-θf)
Hb=900-9θf
Applying the principle of calorimeter
Heat gain = Heat loss
Hc+Hw=Hb
26.4θf-528 + 100θf-2000=900-9θf
26.4θf+100θf+9θf=900+2000+528
135.4θf=3428
Then, θf=3428/133.4
θf=25.32°C
