How much heat would be absorbed by 75.20 g of iron when heated from 22 c to 28 c

Mass x SH x °C (or K) ΔT

= 75g x 0.45J/g/K x 6.0 ΔT

= 202.5 Joules of heat absorbed.

(202.5J / 4.184J/cal = 48.4 calories).

I guess that is the answer

Hello friend ☺

ΔH = MCΔT

ΔH = to the amount of energy or change in energy (J)

M = mass of substance

C = waters specific heat capacity

ΔT = change in temperature

and so ΔH = 75.20 * 4.18 * ( 28-22 ) = 1886.016 J

Thanks ❤

yup its correct its: c) b b

be subject to a force andchange its motion