6. an unknown material has a normal melting/freezing point of 225.0 °c, and the liquid
phase has a specific heat capacity of 160 j/(kg? cº). one-tenth of a kilogram of the solid
at 225.0 c is put into a 0.150-kg aluminum calorimeter cup that contains 0.100 kg ot
glycerin. the temperature of the cup and the glycerin is initially 27.0 °c. all the
unknown material melts, and the final temperature at equilibrium is 20.0 °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?
(5)

19.4 kJ/kg

Explanation:

I'm assuming you meant that the melting/freezing point is -25.0 °C.

Heat gained by the unknown material = heat lost by the oil and aluminum

mL + mCΔT = -(mCΔT + mCΔT)

First, let's find the amount of heat gained by the unknown material:

(0.100 kg) L + (0.100 kg) (160 J/kg/°C) (20.0°C − (-25.0°C))

(0.100 kg) L + 720 J

The heat lost by the oil:

(0.100 kg) (2430 J/kg/°C) (20.0°C − 27.0°C)

-1701 J

The heat lost by the aluminum:

(0.150 kg) (910 J/kg/°C) (20.0°C − 27.0°C)

-955.5 J

Therefore:

(0.100 kg) L + 720 J = -(-1701 J + (-955.5 J))

(0.100 kg) L + 720 J = 2656.5 J

(0.100 kg) L = 1936.5 J

L = 19365 J/kg

L = 19.4 kJ/kg

measure their densities same density, same sub

c because the news are a source of information of things that happen on the daily and shows evidence about that particular topic