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

Answer:

The power required to cool the water is 1.11Kw.

Hence the correct option is (b).

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:

$\rho=\frac{m}{V}$

$1000=\frac{m}{(1l)(\frac{1m^{3}}{1000l})}$

m=1kg

Step2

Amount of heat extraction is calculated as follows:

$Q=mc\bigtriangleup T$

$Q=1(4.186kj/kgk)(\frac{1000 j/kgk}{1 kj/kgk})\times(100-20)$

Q=334880 j.

Step3

Power to cool the water is calculated as follows:

$P=\frac{Q}{t}$

$P=\frac{334880}{(5min)(\frac{60s}{1min})}$

P=1116.26W

or

$P=(1116.26W)(\frac{1Kw}{1000 W})$

P=1.11 Kw.

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

Hence the correct option is (b).