What energy is required to melt a lead bar weighing 0.5 kg, taken at a temperature of 27 degrees Celsius?
| June 28, 2021 | education
m = 0.5 kg.
t1 = 27 ° C.
t2 = 327 ° C.
C = 140 J / kg * ° C.
q = 25 * 10 ^ 3 J / kg.
Q -?
The amount of thermal energy Q will be the sum of: the amount of thermal energy Q1, which is required to heat the bar from t1 to the melting temperature t2, and the amount of thermal energy Q2 required to melt at this temperature.
Q = Q1 + Q2.
Q1 = C * m * (t2 – t1).
Q1 = 140 J / kg * ° C * 0.5 kg * (327 ° C – 27 ° C) = 21000 J.
Q2 = q * m.
Q2 = 25 * 10 ^ 3 J / kg * 0.5 kg = 12500 J.
Q = 21000 J + 12500 J = 33500 J.
Answer: to melt a lead bar, you need Q = 33500 J of thermal energy.
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