An unknown material has a normal melting/freezing point of -26.3 ° C, and the li

Question

An unknown material has a normal melting/freezing point of -26.3 ° C, and the liquid phase has a specific heat capacity of 160 J/(kg · C °). One-tenth of a kilogram of the solid at -26.3 ° C is put into a 0.143-kg aluminum calorimeter cup that contains 0.095 kg of glycerin. The temperature of the cup and the glycerin is initially 28.3 ° C. All the unknown material melts, and the final temperature at equilibrium is 19.6 ° 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?
J/kg

Explanation / Answer

The basic equation is the heat energy equation:

Q = mCpΔT

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

For Al, 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
-25 °C to 20.0 °C is

Qu = mL + mCpΔT

where m = 0.1 kg, L = specific latent heat of fusion, Cp = 160 J/(kg ·C °),
and ΔT = (20.0 - (-25.0)) = 45 °C, u = unknown material designation.

So Qu = 0.1L + 720 J.....(i)

The heat energy loss by glycerine is:

Qg = 0.1 kg * 2430 J/kg- °C * (20 - 27)

or Qg = -1701 J...(ii)

The heat energy loss by the Al calorimeter is

Qal = 0.15 * 910 * (20 - 27)

or Qal = - 955.5 J...(iii)

Heat gained = heat loss

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

Qu = 0.1L + 720 J = Qg + Qal = (- 1701 - 955.5) J

So, 0.1L = -1701 - 955.5 - 720

or 0.1L = -3376.5 J <======Ans

or L = -33765 J/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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