Answer:
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
