How much energy will be released during crystallization and cooling from the melting point

| August 15, 2021 | education

How much energy will be released during crystallization and cooling from the melting point to 27 degrees of a lead plate measuring 2 by 5 by 10 centimeters.

t1 = 27 C
t2 = 327 C – melting point of lead
a = 2 cm = 0.02 m – plate length
b = 5 cm = 0.05 m – plate width
s = 10 cm = 0.1 m – plate thickness
L = 0.25 * 10 ^ 5 J / kg – specific heat of fusion of lead
С = 140 J / kg * С – specific heat of lead
Q -?
Q1 = L * m – the amount of heat released during crystallization of lead
Q2 = C * m * (t2-t1) – the amount of heat released when cooling the lead
Q = Q1 + Q2
m = p * V
p = 11300 kg / m ^ 3 – lead density
V = a * b * c
m = 0.02 * 0.05 * 0.1 * 11300 = 1.13 kg
Q1 = 0.25 * 10 ^ 5 * 1.13 = 28250 J
Q2 = 140 * 1.13 * (327-27) = 47460 J
Q = 28250 +47460 = 75710 J



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