Question

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)

Answer 1

The latent heat of fusion of a substance is the amount of heat energy required to change one kilogram of the substance from a solid to a liquid state at its melting point.

Step-by-step explanation:

The latent heat of fusion of a substance is the amount of heat energy required to change one kilogram of the substance from a solid to a liquid state at its melting point. To calculate the latent heat of fusion, we can use the equation:

Q = m * Lf

where Q is the heat energy absorbed, m is the mass of the substance, and Lf is the latent heat of fusion. In this case, we know the mass of the unknown material (0.1 kg) and the temperature change (the difference between the initial temperature of the solid and the final temperature of the equilibrium). Using the given values, we can solve for the latent heat of fusion.

Answer 2

19.4 kJ/kg

Step-by-step 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