An unknown material has a normal melting/freezing point of -25.3 °C, and the liq

Question

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?

Explanation / Answer

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.

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