Question

An unknown material has a normal melting/freezing point of -24.4 °C, and the liquid phase has a specific heat capacity of 160 J/(kg C°). One-tenth of a kilogram of the solid at -24.4 °C is put into a 0.157-kg aluminum calorimeter cup that contains 0.198 kg of glycerin. The temperature of the cup and the glycerin is initially 26.1 °C. All the unknown material melts, and the final temperature at equilibrium is 19.9 °C. The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

Answer 1

Write the basic expression of the heat energy balance -  

Q = mCp?T

where m is the mass of the substance, Cp is its specific heat capacity
and ?T is the change in temperature

Now for aluminium, I will use Cp(al) = 910 J/kg-°C

For glycerine, I will use Cp(gly) = 2430 J/kg-°C

The heat energy to raise the temp of the unknown material from
-24.4°C to 19.9°C is

Qu = mL + mCp?T

where m = 0.198 kg, L = specific latent heat of fusion, Cp = 160 J/(kg·C°),
and ?T = (19.9 - (-24.4)) = 44.3°C, u = unknown material designation.

So Qu = 0.1L + 0.1*160*44.3 = 0.1L + 708 J -----------------------(i)

The heat energy loss by glycerine is:

Qg = 0.198 kg * 2430 J/kg-°C * (19.9 - 26.1)

=> Qg = -2983 J ---------------------------------------(ii)

The heat energy loss by the Al calorimeter is

Qal = 0.157* 910 * (19.9 - 26.1)

or Qal = - 886 J----------------------------------------------(iii)

Heat gained = heat loss

From (i), (ii) and (iii) we get

Qu = 0.1L + 708 J = Qg + Qal = (- 2983 - 886) J

=> 0.1L = -2983 - 886 - 708

=> 0.1L = -4577 J

=> L = - 45770 J/kg

Therefore, the latent heat of fusion of the unknown material = 45770 J/kg.