Question

Please explain in detail.

a) A student performed a calorimetry experiment. He combined 50.0 mL of water at 85˚C and 50.0 mL of water at 22˚C in a coffee cup calorimeter. The final temperature of the water was 45˚C. What was the heat capacity of the calorimeter?

b) The same calorimeter was used to find the specific heat of a metal. He heated 95.22 grams of the metal in a boiling water bath (99˚C). After adding this hot metal to 100.0 mL of 21˚C water in the calorimeter, the final temperature reached 29˚C. What is the metal’s specific heat?

Answer 1

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

Or,

            -Q_H (hot water) = Qw (cold water) + Qc (Calorimeter)

            Or, -m₁ s (dT) = m₂ s (dT) + C x dT

Where,

q = heat gained or lost

m = mass of water (hot or cold)

s = specific heat of water

dT = Final temperature – Initial temperature

C = Heat capacity of calorimeter

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

It’s assumed that the density of water is equal to 1.000 g/mL so that the volume of water in mL is numerically equal to its respective mass in grams.

The –ve sign of Q_H indicates that hot water loses heat when mixed with cold water.

Now,

-50.0 g x 4.184 J g^{-1 0}C^-1 (45- 85)⁰C =

[50.0 g x 4.184 J g^{-1 0}C^-1 (45- 22)⁰C] + [ C x (45-22)⁰C]

            Or, 8368 J = 4811.6 J + C x (23.0⁰C)

            Or, C x (23.0⁰C) = 8368 J - 4811.6 J

            Or, C = 3556.4 J / (23.0⁰C)

            Hence, C = 154.63 J ⁰C^-1

Therefore, heat capacity of calorimeter, C = 154.63 J ⁰C^-1

#B. The total amount of heat lost by hot metal is equal to the sum of heat gained by- I. Cold water, and II. Calorimeter for the establishment of thermal equilibrium.

Or,

            -Q_m (hot metal) = Qw (cold water) + Qc (Calorimeter)

            Or, -m₁ s_m (dT) = m₂ s (dT) + C x dT

Where, s_m = specific heat of metal

Or, -95.22 g x s_m (29-99)⁰C =

[100.0 g x 4.184 J g^{-1 0}C^-1 (29-21)⁰C] + [ 154.63 J ⁰C^-1 x (29-21)⁰C]

            Or, s_m x (6665.4 g ⁰C) = 3347.2 J + 1237.04 J

            Or, s_m x (6665.4 g ⁰C) = 4584.24 J

            Or, s_m = 4584.24 J / (6665.4 g ⁰C) =

            Hence, s_m = 0.688 J g^-10C^-1

Therefore, specific heat of metal, s_m = 0.688 J g^-10C^-1