A mixture of 125 g of iron and 250 g of water is heated from 25.0 °C to 70.0 °C.How much heat is absorbed by the mixture? (Specific heat of water = 4.18 J/g °C;iron = 0.473 J/g °C)

Question

Quizplus · Accepted Answer

First we need to calculate the heat absorbed by the water and then the iron. Using the formula Q = mcΔT, where Q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature:

For water:
Q = (250 g)(4.18 J/g °C)(45.0 °C) = 46950 J = 46.9 kJ (rounded to 3 significant figures)

For iron:
Q = (125 g)(0.473 J/g °C)(45.0 °C) = 2672 J = 2.67 kJ (rounded to 3 significant figures)

Total heat absorbed:
Q_total = 46.9 kJ + 2.67 kJ ≈ 49.7 kJ

A Mixture of 125 G of Iron and 250 G