Question

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

Answer 1

Answer:

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

Answer 2

Answer:

2667 kj/kg

Explanation:

After 120 seconds the electric kettle would have produced a total heat of 2000J * 120 = 240000J or 240 kj

Since it's 90 g, or 0.09 kg lighter, this means the 240 kj is used to transform liquid water to steam. So the specific latent heat would be

240 / 0.09 = 2667 kj/kg