Answer:
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)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/c2qts5thhng6fcq2l92p1lw8vy29dwoyfj.png)
![1000=(m)/((1l)((1m^(3))/(1000l)))](https://img.qammunity.org/2020/formulas/engineering/college/eeq588u1miu1t24a53mozai1rqspr42zyp.png)
m=1kg
Step2
Amount of heat extraction is calculated as follows:
![Q=mc\bigtriangleup T](https://img.qammunity.org/2020/formulas/engineering/college/scy41xhm1zfyhiv5mz3kqg6c4p1aq1h5dv.png)
![Q=1(4.186kj/kgk)((1000 j/kgk)/(1 kj/kgk))*(100-20)](https://img.qammunity.org/2020/formulas/engineering/college/umrqy67d20vqepg9zarwbktvstfvqa4yfw.png)
Q=334880 j.
Step3
Power to cool the water is calculated as follows:
![P=(Q)/(t)](https://img.qammunity.org/2020/formulas/engineering/college/yvdoz5nrj8iuie7a5we71pqyqr54nvhzci.png)
![P=(334880)/((5min)((60s)/(1min)))](https://img.qammunity.org/2020/formulas/engineering/college/8lip4n1jpq0ked6uplb9egn8okr18vbp6z.png)
P=1116.26W
or
![P=(1116.26W)((1Kw)/(1000 W))](https://img.qammunity.org/2020/formulas/engineering/college/f87o6p4r6wu9vyqzw2bfb34z4s0j3dyjva.png)
P=1.11 Kw.
Thus, the power required to cool the water is 1.11Kw.
Hence the correct option is (b).