Question

1. Two liters of methanol at -30 ˚C and two liters of water at 25 ˚C are mixed in an

insulated glass cylinder that has a mass of 2 kg. The heat capacity of glass is 0.9 kJ/kgK.Using the properties of the other substance in the appendix of the book, determine thefinal temperature of the mixture.

at room temp

Answer 1

Ans. when methanol (-30⁰C) is mixed with water (25⁰C), some heat will be exchanged between the two liquids. However, some heat will also be exchanged with the glass cylinder because it has accountable specific heat.

We will proceed in two steps:

Step 1: Methanol is mixed with water.

Heat change (q) of a system (substance) is given by-

q = m x s x dT                      - equation 1   

Where,

                                                m = mass in kg

s = specific heat of the substance, kJ/ (Kg⁰C)

dT = change in temperature = (final – initial) temperature

Let the equilibrium temperature of the mixture be T.

Heat gained by methanol (as its temperature shall increase from -30⁰C towards that of water) must be equal to heat lost by water.

Mass of 2.0 L methanol at -30⁰C = Volume x density

                                                = 2.0 L x (0.838 kg/L) = 1.676 kg

            Or, q (methanol) = - q (water)            ; -ve sign indicates heat loss by water

            Or, [ 1.676 kg x (2.48 kJ / kg⁰C) x {T- (-30⁰C)]} = - [2.0 kg x (4.184 kJ / kg⁰C) x (T- 25⁰C)]

            Or, (4.15648 kJ/⁰C) x (T + 30⁰C) = (8.368 kJ/ ⁰C) x (25⁰C- T)

            Or, (4.15648 kJ/⁰C) / (8.368 kJ/ ⁰C) = (25⁰C- T) / (T + 30⁰C)

            Or, 0.50 x (T + 30⁰C) = (25⁰C- T)

            Or, 0.5T + 15⁰C = 25⁰C – T

            Or, 1.5 T = 10⁰C

            Or, T = 10⁰C/ 1.5 = 6.67⁰C

Hence, equilibrium temperature of the methanol-water mixture = 6.67⁰C

Step 2:

From step 1-

            I. Total mass of methanol-water mixture = 1.676 kg + 2.0 kg = 3.676 kg

            II. Average specific heat of the mixture = [(2.48 + 4.184)/2] kJ / kg⁰C = 3.332 kJ / kg⁰C

Note: It is not the exact value, the calculation has been made assuming both the components have equal mass. You can proceed with assuming the specific heat of the mixture equal to that of water. But since both the components are nearly equal in mass, the later assumption may cause greater deviation from the practical values.

            III. Mass of glass cylinder = 2.0 kg

                        Temperature = room temperature = 25⁰C

                        Specific heat = 0.90 kJ / kg⁰C

Now, we have two system that are about to attain thermal equilibrium. Let the equilibrium temperature by Y.

Again, methanol-water mixture (at 6.67⁰C) gains heat and glass cylinder (at 25⁰C) loses heat.

Putting the values as in step 1-

            3.676 kg x (3.332 kJ / kg⁰C) x (Y – 6.67⁰C) = - 2.0 kg x (0.9 kJ / kg⁰C) x (Y – 25⁰C)

`           Or, (12.248432 kJ / ⁰C) / (1.8 kJ / ⁰C) = (25⁰C- Y) / (Y – 6.67⁰C)

            Or, 6.8 x (Y – 6.67⁰C) = (25⁰C- Y)

            Or, 6.8Y + Y = 25⁰C + 45.356⁰C

            Or, Y = 70.356⁰C / 6.8 = 10.34            ⁰C

            Hence, Y = 10.34⁰C

Thus, final equilibrium temperature of the methanol-water-glass cylinder system =10.34⁰C