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:
[tex]\rho=\frac{m}{V}[/tex]
[tex]1000=\frac{m}{(1l)(\frac{1m^{3}}{1000l})}[/tex]
m=1kg
Step2
Amount of heat extraction is calculated as follows:
[tex]Q=mc\bigtriangleup T[/tex]
[tex]Q=1(4.186kj/kgk)(\frac{1000 j/kgk}{1 kj/kgk})\times(100-20)[/tex]
Q=334880 j.
Step3
Power to cool the water is calculated as follows:
[tex]P=\frac{Q}{t}[/tex]
[tex]P=\frac{334880}{(5min)(\frac{60s}{1min})}[/tex]
P=1116.26W
or
[tex]P=(1116.26W)(\frac{1Kw}{1000 W})[/tex]
P=1.11 Kw.
Thus, the power required to cool the water is 1.11Kw.
Hence the correct option is (b).
