How much energy does it take to melt 300 grams of aluminum t1 = 20 degrees

m = 300 g = 0.3 kg.

t1 = 20 ° C.

t2 = 660 ° C.

C = 920 J / kg * ° C.

λ = 3.9 * 10 ^ 5 J / kg.

Q -?

The required amount of heat Q will be the sum: Q = Q1 + Q2, where Q1 is the amount of heat that is needed to heat aluminum from t1 to the melting point t2, Q2 is the amount of heat that is needed to rule it at this temperature.

The amount of heat Q1 required for heating is determined by the formula: Q = C * m * (t2 – t1), where C is the specific heat capacity of aluminum, m is the mass of aluminum, t2, t1 are the final and initial temperatures.

The amount of heat Q2, which is necessary for melting, is determined by the formula: Q = λ * m, where λ is the specific heat of fusion, m is the mass of aluminum.

Q = 920 J / kg * ° C * 0.3 kg * (660 ° C – 20 ° C) + 3.9 * 10 ^ 5 J / kg * 0.3 kg = 293640 J.

Answer: for heating and melting aluminum, Q = 293640 J of thermal energy is needed.