Iron heated to a temperature of 540 C was immersed in water weighing 2 kg, taken at a temperature of 10 C.

Iron heated to a temperature of 540 C was immersed in water weighing 2 kg, taken at a temperature of 10 C. Determine the mass of iron if the steady-state temperature became equal to 40 C.

The amount of heat that will be transferred to the water:
Q = С1 * m1 * ∆t1, С1 = 4200 J / (kg * K), m1 = 2 kg, ∆t1 = tк – tн, tк = 40 ºС, tн = 10 ºС.
The amount of heat that iron will give:
Q = С2 * m2 * ∆t2, where С2 = 460 J / (kg * K)), ∆t2 = tн – tк, tн = 540 ºС, tн = 40 ºС.
Let us express and calculate the desired mass of iron:
C1 * m1 * ∆t1 = C2 * m2 * ∆t2.
m2 = C1 * m1 * ∆t1 / (C2 * m2 * ∆t2) = 4200 * 2 * (40 – 10) / (460 * (540 – 40)) = 252000/2300000 ≈ 1.1 kg.
Answer: The mass of iron is 1.1 kg.