An unknown material has a normal meltingfreezing point of 25

An unknown material has a normal melting/freezing point of -25.3 °C, and the liquid phase has a specific heat capacity of 160 J/(kg C°). One-tenth of a kilogram of the solid at -25.3 °C is put into a 0.130-kg aluminum calorimeter cup that contains 0.132 kg of glycerin. The temperature of the cup and the glycerin is initially 27.8 °C. All the unknown material melts, and the final temperature at equilibrium is 18.2 °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?

Solution

The basic equation is the heat energy equation:

Q = mCpT

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

For Al, Cp(al) = 910 J/kg-°C

For glycerine, Cp(gly) = 2430 J/kg-°C

The heat energy to raise the temp of the unknown material from
-25.3°C to 18.2°C is

Qu = mL + mCpT

where m = 0.132 kg, L = specific latent heat of fusion, Cp = 160 J/(kg·C°),
and T = (18.2 - (-25.3)) = 43.5°C, u = unknown material designation.

So Qu = 0.132L + 918.72 J.....(i)

The heat energy loss by glycerine is:

Qg = 0.132 kg * 2430 J/kg-°C * (18.2 - 27.8)

or Qg = -3079.3 J...(ii)

The heat energy loss by the Al calorimeter is

Qal = 0.13 * 910 * (18.2 - 27.8)

or Qal = - 1135.7 J...(iii)

Heat gained = heat loss

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

Qu = 0.132L + 918.72 J = Qg + Qal = (-3079.3 - 1135.7) J

So, 0.132L = -3079.3 - 1135.7 - 918.72

or 0.132L = -5133.72 J <======Ans

or L = -38891.82J/kg where L is the specific latent heat

The negative sign says you must supply that heat to change the material
to a liquid state.