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  • Please,please help me Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 270C to 770C ? (Specific h

 Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 270C to 770C ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

 

  • Option 1)

    8 min 20 s

  • Option 2)

    6 min 2s

  • Option 3)

    7 min

  • Option 4)

    14 min

 

Answers (1)

best_answer

As we have learned

Power if the force is variable -

P_{av}= \frac{\Delta w}{\Delta t}= \frac{\int_{0}^{t}p\cdot dt}{\int_{0}^{t}dt}

- wherein

P\rightarrow power

dt\rightarrow short\: interval \: of\: time

 

 Let temprature of water rises from 27 \degree C \: \: to \: \: 77 \degree C  in  t second 

Net Heat absorbed in t sec = (1000-160)t 

= 840 J 

Q = ms \Delta T \: \: or \: \: 840 t = (2Kg )\times 4200\times 50

t = \frac{840 \times 500}{840}sec =500 sec = 8 min 20 sec

 

 

 

 

 

 


Option 1)

8 min 20 s

Option 2)

6 min 2s

Option 3)

7 min

Option 4)

14 min

Posted by

SudhirSol

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