A blacksmith heats a 1,540 g iron horseshoe to a temperature of 1445°c before dropping it into 4,280 g of water at 23.1°c. if the specific heat of iron is 0.4494 j / g °c, and the water absorbs 947,000 j of energy from the horseshoe, what is the final temperature of the horseshoe-water system after mixing

Given:

m₁ = 1540 g, mass of iron horseshoe

T₁ = 1445 °C, initial temperature of horseshoe

c₁ = 0.4494 J / (g-°C), specific heat

m₂ = 4280 g, mass of water

T₂ = 23.1 C, initial temperature of water

c₂ = 4.18 J / (g-°C), specific heat of water

L = 947,000 J heat absorbed by the water.

Let the final temperature be T °C.

For energy balance,

m₁c₁ (T₁ - T) = m₂c₂ (T - T₂) + L

(1540 g) * (0.4494 J / (g-C)) * (1445-T C) = (4280 g) * (4.18 J / (g-C)) * (T-23.1 C) + 947000 J

692.076 (1445 - T) = 17890 (T - 23.1) + 947000

10⁶ - 692.076T = 17890T - 413259 + 947000

466259 = 18582.076T

T = 25.09 °C

Answer: 25.1 °C