A. If 0.45 kg of iron at 140°C is added to 1.5 kg of water at 22°C, what will be the final temperature of the system? (4 points)

25.67 °C

Explanation:

From the question,

Heat lost by iron = heat gained by water

c'm'(t₁-t₃) = cm(t₃-t₂) Equation 1

Where c' = specific heat capacity of iron, m' = mass of iron, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature of iron, t₂ = initial temperature of water, t₃ = temperature of the system.

make t₃ the subject of the equation

t₃ = (c'm't₁+cmt₂)/(c'm'+cm) Equation 2

Given: m' = 0.45 kg, t₁ = 140 °C, t₂ = 22 °C, m = 1.5 kg

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

Substitute these values into equation 2

t₃ = (0.45×450×140+1.5×4200×22)/(450×0.45+4200×1.5)

t₃ = (28350+138600)/((202.5+6300)

t₃ = 166950/6502.5

t₃ = 25.67 °C

i think if it were me i whould divide 300 dide 15

step-by-step explanation: