Question

7. A piece of zinc metal with a mass of 35.0 g is heated in boiling water to 101.3 ºC and then dropped into a coffee-cup calorimeter containing 75.0 g of water at 23.8 ºC. When thermal equilibrium is reached, the final temperature is 27.0 ºC. Calculate the specific heat capacity of the zinc.

Answer 1

Using, q = m c dT       -- equation 1   

Where, q= heat change                                                                  , m= mass in gram,

C= specific heat or heat capacity (in terms of J/g⁰C)                           

dT = Final – Initial temperature

Part 1: Heat gained by Zn metal

                Let, Specific heat of Zn = c1

Initial temperature of Zn metal = T1

Given, mass of Zn = 35.0 g

                Now, q1 = 35.0 g x c1 x (101.3 – T1)         

Part 2: When heated Zn metal is placed in calorimeter, the heat lost by Zn is absorbed by water to increase its temperature.

            Now,

            Specific heat of water = c2 = 4.18 J/ g⁰C

Initial temperature of water = 23.8⁰C

Final temperature of water = 27.0⁰C

dT2 = (27.0 – 23.8)⁰C = 3.2⁰C

Mass of water = 75.0 g

                Now, q2 = 75.0 g x (4.18 J/ g⁰C) x 3.2⁰C = 1003.2 J

Part 3: heat lost by Zn metal is calorimeter is gained by water.

Therefore, - q1 = q2      - equation 2

Note: the -ve sign of -q1 indicates that heat is released from Zn metal.

Now,

            q1 = q2                        [ignoring the -ve sign, it does not affect the result.]

            or, 35.0 g x c1 x (101.3 – T1)⁰C = 1003.2 J

                or, c1 = 1003.2 J / [35.0 g x (101.3 – T1)⁰C]

Note 2: The initial temperature of Zn is missing from question. Put the value of T1 in above equation to get the value of c1.