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T = 40.08 °C

Explanation:

Given:

m₁ = 0.33 kg, mass of iron horseshoe

T₁ = 693°C, initial temperature of horseshoe

c₁ = 448 J/(g-°C), specific heat

m₂ = 24 kg, mass of water

T₂ = 39°C, initial temperature of water

c₂ = 4186 J/(g-°C), specific heat of water

let T be the equilibrium temperature.

For energy balance,

m₁c₁(T₁ - T) = m₂c₂(T - T₂)

(0.330) * (448) * (693 - T) = (24) * (4186) * (T - 39)

147.84 * (693 - T) = 100,464 * (T - 39)

102,453.12 - 147.84T = 100,464T - 3,918,096

102,453.12 + 3,918,096 = 100,464T - 147.84T

4020549.12 = 100316.16T

T = 40.08 °C