Question

how many calories of energy are given off to lower the temperature of 100.0g of iron from 150.0°C to 35.0°C?

Answer 1

ΔT = Tfinal − Tinitial = 150°C − 35.0°C = 125°C

given the specific heat of iron as 0.108 cal/g·°C

heat=(100.0 g)(0.108 cal /g· °C )(125°C) =

100x 0.108x125= 1350 cal