How much heat would be needed to completely evaporate 31.5 g of boiling water at a temperature of 100 "c? express your answer in units of joules.

Heat needed = 71.19 J

Explanation:

Here heat required can be calculated by the formula

            H = mL

M is the mass of water and L is the latent heat of vaporization.

Mass of water, m = 31.5 g = 0.0315 kg

Latent heat of vaporization of water = 2260 kJ/kg

Substituting

            H = mL = 0.0315 x 2260 = 71.19 kJ

Heat needed = 71.19 J

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.