Question

How much power would you need to cool down a closed, 1 Liter container of water from 100°C to 20°C in 5 minutes? (a) 1.1W (b)1.1kW (c)67kW (d)334 kJ

Answer 1

The power required to cool the water is 1.11Kw.

Hence the correct option is (b).

Step-by-step explanation:

Power needed to cool down is equal to heat extract from the water.

Given:

Volume of water is 1 liter.

Initial temperature is 100C.

Final temperature is 20C.

Time is 5 minutes.

Take density of water as 100 kg/m3.

Specific heat of water is 4.186 kj/kgK.

Calculation:

Step1

Mass of the water is calculated as follows:

`\rho=(m)/(V)`

`1000=(m)/((1l)((1m^(3))/(1000l)))`

m=1kg

Step2

Amount of heat extraction is calculated as follows:

`Q=mc\bigtriangleup T`

`Q=1(4.186kj/kgk)((1000 j/kgk)/(1 kj/kgk))*(100-20)`

Q=334880 j.

Step3

Power to cool the water is calculated as follows:

`P=(Q)/(t)`

`P=(334880)/((5min)((60s)/(1min)))`

P=1116.26W

or

`P=(1116.26W)((1Kw)/(1000 W))`

P=1.11 Kw.

Thus, the power required to cool the water is 1.11Kw.

Hence the correct option is (b).