An aluminum electric tea kettle with a mass of 500 g is heated with a 500-w heating coil. how long will it take to heat up 1.0 kg of water from 18°c to 98°c in the tea kettle

669.76s or 11.16mins

Explanation:

Mass of kettle (Mk) = 500g

Mass of water (Mw) = 1.0kg = 1000g

Power = 500w

θ₁ = 18°C

θ₂ = 98°C

specific heat capacity of water (c) = 4.186J/g*°C

Q (work done) = ?

Q = MC∇θ

Q = Mc (θ₂ - θ₁)

The work done in raising 1kg of water from 18°c to 98°c = Q

Q = 1000 * 4.186 * (98 - 18)

Q = 1000 * 4.186 * 80

Q = 334880J

But power = workdone / time

Time = workdone / power

Time = Q / P

Time = 334880 / 500

Time = 669.76s

T = 11.16mins

The time taken to heat 1kg of water from 18°C to 98°C is 669.76s

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min