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.
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