Question

A 465 g chunk of iron is removed from an oven and plunged into 375 g water in an insulated container. The temperature of the water increases from 26.0 to 87.0 oC. If the specific heat of iron is 0.449 J/g-oC, what must have been the original temperature of the iron?

Answer 1

heat lost by the iron = heat gained by the water

heat gained by the water (q) $= m \times S \times ΔT$

$= 375 \times 4.184 \times (87 – 26) \\ = 95709\; J$

heat lost by the iron = 95709 J

$-95709 = 465 \times 0.449 \times ( 87 -T_1) \\ T_1 = 545.41\; °C$

original temperature of the iron is 545.41 °C