Question

Not really sure how to go about solving for the specific heat of the steel cube.
My data is written in the table. Given value for calorimeter cup was 0.27

Law of Heat Exchange

Test specimen	Wt. empty cup	Wt. cup w/ water	Wt. water	Wt. metal	Tc	Th	Tf	Specific heat
Steel cube	3.13 g	124.83 g	121.7 g	257.93g	22.4	85.8	29	?

Basically for the procedure we first heated a steel cube in boiling water. Then filled a calorimeter cup halfway with water and recorded the temperature of the water (Tc). Then after 5 minutes, recorded temperature of hot water (Th), then put cube in cup with cold water, stirred, and recorded final water temp (Tf).

Answer 1

Ans. The total amount of heat lost by steel cube is equal to the sum of heat gained by- I. Cold water, and II. Calorimeter for the establishment of thermal equilibrium.

Or,

            -Q_sc (steel cube) = Qw (cold water) + Qc (Calorimeter)

            Or, -m₁ s_s (dT₁) = m₂ s (dT₂) + m₃ x s_c (dT_­2)

Where,

q = heat gained or lost

m = mass of water (hot or cold)

s = specific heat of water

s_s = specific heat of steel

s_c = specific heat of calorimeter cup

dT = Final temperature – Initial temperature

It’s assumed that the initial temperature of calorimeter is equal to the initial temperature of cold water put in it.

The –ve sign of Q_sc indicates that steel cube loses heat when put in calorimeter.

It’s assumed that the “given value for calorimeter cup was 0.27” REFERS to the specific heat of calorimeter cup.

Now,

Or, -257.93 g x s_s (29 – 85.8)⁰C =

[121.7 g x 4.184 J g^{-1 0}C^-1 (29-22.4)⁰C] + [ 3.13 g x 0.27 J g^{-1 0}C^-1 x (29-22.4)⁰C]

            Or, s_s x (14650.424 g ⁰C) = 3360.67248 J + 5.57766 J

            Or, s_s x (14650.424 g ⁰C) = 3366.25014 J

            Or, s_s = 3366.25014 J / (14650.424 g ⁰C)

            Hence, s_s = 0.23 J g^-10C^-1

Therefore, specific heat of steel cube = 0.23 J g^-10C^-1