10. If 3.5 kJ of energy are added to a 28.2 g sample of iron at 20°C, what is the final temperature of the iron in kelvins? The specific heat of iron
is 0.449 J(g•K).

569K

Explanation:

Q = 3.5kJ = 3500J

mass = 28.2g

∅1 = 20°C = 20 + 273 = 293K

∅2 = x

c = 0.449

Q = mc∆∅

3500 = 28.2×0.449×∆∅

3500 = 12.6618×∆∅

∆∅ = 3500/12.6618

∆∅ = 276.4220

∅2 - ∅1 = 276.4220

∅2 = 276.4220 + ∅1

∅2 = 276.4220 + 293

∅2 = 569.4220K

∅2 = 569K

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.