Assumptions :

Specific heat capacity of water = 4.2 kJ / kg C° for water , 0.45 kJ / kg C° for iron

Density of water = 1kg/L (or 1 g/ml, whichever you prefer)

Mass of 1.05 L of water = 1.05 kg


Energy needed to raise temperature of the 1.05 L of water and 0.3kg iron pot from 20.0 °C to 25.0 °C

= mc ∆θ (where m is mass, c is specific heat capacity and ∆θ is the change in temperature)

= (1.05 kg)(4.2 kJ / kg C°)(25.0 °C - 20.0 °C) + (0.3 kg)(0.45 kJ / kg C°)(25.0 °C - 20.0 °C)

= 22.05 kJ + 0.675 kJ

= 22.725 kJ

q = mc ∆θ

22.725 kJ of energy was lost by the horseshoe to the water and pot.

Put a negative sign since energy loss is represented as a negative value.

-22.725 kJ = (0.4 kg)(0.45 kJ / kg C°)(∆θ)

∆θ = -22.725 kJ ÷ 0.4 kg ÷ 0.45 kJ / kg C°

∆θ = -126.25 °C


So the temperature of the horseshoe decreased by 126.25 °C. The horseshoe is now at the equilibrium temperature of 25.0 °C.

Initial temperature of horseshoe

= 25.0 °C + 126.25 °C

= 151.25 °C


Your answer will vary depending on the value of c you use for water and iron, as well as the value you use for the density of water.