4. An electric kettle produces 2000 J of energy each second. It is filled with water, weighed and switched on. After coming to the boil, it is left on for a further 120 seconds and is then switched off. It is found to be 90 g lighter. Calculate the specific latent heat of vaporisation from this data

2.67 × 10⁶ J/kg

Explanation:

Since the electric kettle is 90 g lighter, this means that 90 g of water evaporates.

Now, Q = mL where Q = quantity of heat = Pt where P = energy per second of kettle = 2000 J/s and t = time = 120 s. m = mass of water evaporated = 90 g = 0.090 g and L = specific latent heat of vaporisation of water.

L = Q/m = Pt/m = 2000 J/s × 120 s/ 0.090 kg = 2.67 × 10⁶ J/kg

explanation:

the masses of the objects and the distance between them

uranium fissions/breaks up giving out energy and neutrons