To what temperature should the aluminum cube be heated so that it, being put on ice, is completely immersed in it?

t = 0 “C. ρal = 2700 kg / m ^ 3. ρ ice = 900 kg / m ^ 3. λ ice = 3.4 * 10 ^ 5 J / kg. Sal = 920 J / kn *” C. tal -?

Let’s write down the heat balance equation: Sal * mal * (tal – t) = λ ice * m ice. mal = ρal * V, m ice = ρ ice * V. Sal * ρal * V * (tal – t) = λ ice * ρ ice * V. Sal * ρal * tal – Sal * ρal * t = λ ice * ρ ice. Sal * ρal * tal = λ ice * ρ ice + Sal * ρal * t. tal = (λ ice * ρ ice + Sal * ρal * t) / Sal * ρal. Since t = 0 “С, the formula will take the form: tal = λled * ρled / Sal * ρal. Tal = 3.4 * 10 ^ 5 J / kg * 900 kg / m ^ 3/920 J / kn *” C * 2700 kg / m ^ 3 = 123 “C.
Answer: the aluminum cube must be heated to t = 123 “C.