Question

calorimeter has aluminum inner cup of mass 120 gram containing 100 ml water at temperature 20 degree Celsius. Brass piece with mass 100 gram is heated to 100 degree Celsius, and then immersed in the calorimeter. Calculate the final temperature of the system. The specific heat of brass is 0.09 cal/(gramXdeg.C) . The additional necessary data are provided in the text.

Answer 1

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