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

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27