**1 kg of water is contained in a 1.25 kW kettle. Assuming specific heat capacity of water … water which evaporates per minute from the…**

###### Answers:

**(i)** Heat required for the temperature of 1 kg (= 1000 g) of water to rise from 25^{0}C to its boiling point (i.e., 100^{0}C) = Mass x Specific heat capacity x Rise in temperature

= 1000 x 4.2 x (100 – 25) = 31500 J

Power supplied by the kettle = 1.25 kW = 1.25 x 1000 W = 1250 W

Since Power = Energy/Time, ∴ 1250 = 31500/t

Hence time taken t = 315000/1250 = 252 s (or 4 min 12 s)

**(ii)** Energy supplied by the kettle in 1 minute (= 60s) = Power x Time

= 1250 x 60 = 75000 J

Heat required for boiling water to evaporate

= Mass x Specific latent heat of vaporisation

= m x 2260 = 2260 mJ

Thus 2260 m = 75000

Or Mass of water evaporated per minute, m = 75000/2260 = 33.18 g

33.18 g is the answer