50 kg of liquid steel is poured from the ladle into the mold. What energy will be released when it solidifies

50 kg of liquid steel is poured from the ladle into the mold. What energy will be released when it solidifies and cools from the melting point of 1500 degrees Celsius to 0 degrees Celsius.

Steel, mass m, being in a liquid state at the melting temperature, while cooling down, will take part in two processes: solidification and cooling. When solidified, it will release the amount of heat Q₁ = λ ∙ m, where λ is the specific heat of fusion. For steel λ = 84000 J / kg. When solidified, it will release the amount of heat Q₂ = c ∙ m ∙ (t₂ – t₁), where c = 500 J / (kg ∙ ° C) is the specific heat capacity of steel, t₂ and t₁ are the initial and final temperatures of the metal. Total heat will be released:

Q = Q₁ + Q₂ or Q = λ ∙ m + с ∙ m ∙ (t₂ – t₁).

From the condition of the problem it is known that m = 50 kg of liquid steel was poured from the ladle into the mold at a melting temperature t₁ = 1500 ° C, which was cooled to a temperature of t₂ = 0 ° C. Substitute the values ​​of the quantities into the calculation formula and find what energy will be released during its solidification and cooling:

Q = 84000 J / kg ∙ 50 kg + 500 J / (kg ∙ ° С) ∙ 50 kg ∙ | 0 ° С – 1500 ° С |;

Q = 41,700,000 J = 41.7 MJ.

Answer: 41.7 MJ of thermal energy will be released.