What is the minimum number of kiloJoules needed to change 40.0 grams of water at 100C to steam at the same temperature and pressure

The minimum number of kilo Joules needed is 90.4KJ.

Explanation:

The energy required= latent heat of vaporization.

Latent heat of vaporization of water= 2260 KJ/Kg

Mass of water = 0.04 Kg

Heat required= 0.04Kg× 2260 KJ/Kg

Heat required= 90.4KJ.

Learn more: link

90.4Kg

Explanation:

The energy required= latent heat of vaporization

Latent heat of vaporization of water= 2260 KJ/Kg

Mass of water = 0.04 Kg

Heat required= 0.04Kg× 2260 KJ/Kg

Heat required= 90.4Kg

1. they vary with a higher number because boiling is way hotter than melting

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml