Question

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

Answer 1

669.76s or 11.16mins

Step-by-step 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

Answer 2

Time Needed is 12 minutes

Step-by-step explanation:

mass=500g convert it into Kg is 0.5 Kg

Q=mcΔT

Q is heat

c is specific heat constant

m is mass

ΔT is temperature change

First Aluminum Heat

`Q=900*0.5*80\\Q=2700J`

Specfic heat of water

`1(cal)/(g)=4.18(j)/(g)\\  or\\=4180(j)/(kg)`

For Water Heat

`Q=4180*1*80\\Q=334400J`

Total Heat=Aluminum Heat + Water Heat

Q=334400+27000

Q=361400J

Time Needed

`T=(Q)/(W)\\ T=(361400)/(500)\\ T=722.8 seconds\\T=12.04 minutes`