A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 14.0 kg of water at 24.0°C. What is the final temperature of the water–horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away.

29.97 °C

Explanation:

From the question,

Heat lost by the iron horseshoe = heat gained by water.

CM(t₁-t₃) = cm(t₃-t₂) Equation 1

Where C = specific heat capacity of the iron horseshoe, M = mass of the iron horse shoe, c = specific heat capacity of water, m = mass of water, t₁ = Initial temperature of the iron horseshoe, t₂ = initial temperature of water, t₃ = final Temperature of the mixture.

make t₃ the subject of the equation,

t₃ = (CMt₁+cmt₂)/(CM+cm) Equation 2

Given: M = 1.5 kg, m = 14 kg, t₁ = 550 °C, t₂ = 24 °C

Constant: C = 450 J/kg.K, c = 4200 J/kg.K

Substitute into equation 2

t₃ =[(450×1.5×550)+(4200×14×24)]/(450×1.5+4200×14)

t₃ = (371250+1411200)/(675+58800)

t₃ = 1782450/59475

t₃ = 29.97 °C

Hence the final temperature of water = 29.97 °C

The final temperature of the water–horseshoe system is 30⁰C

Explanation:

Given;

mass of iron, = 1.50-kg

initial temperature of the iron horseshoe, = 550°C

mass of water, Mw =  14.0 kg

initial temperature of water,Tw = 24°C

Let the final temperature of the water–horseshoe system = T

From the principles of conservation of heat;

Heat lost by a hot body = Heat gained by a cold body

Heat lost by hot iron horseshoe = heat gained by water

where;

is the specific heat capacity of iron = 449 J/kg.°C

is the specific heat capacity of water = 4200 J/kg°C

Therefore, the final temperature of the water–horseshoe system is 30⁰C