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Question: Water of volume 2 litre in a container is heated with a coil of 1 kW at 27<sup>0</sup>C. The lid of ...

Water of volume 2 litre in a container is heated with a coil of 1 kW at 270C. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 270C to 770C ? [Given specific heat of water is 4.2 kJ/kg]

A

8 min 20 s

B

6 min 2 s

C

7 min

D

14 min

Answer

8 min 20 s

Explanation

Solution

Energy gained by water (in 1 s)

= energy supplied – energy lost

= (1000 J – 160 J) = 840 J

Total heat required to raise the temperature of water from 270 to 770C is msDq.

Hence, the required time

t = msΔθratebywhichenergyisgainedbywater\frac{ms\Delta\theta}{ratebywhichenergyisgainedbywater}

=2×(4.2×103)×50840\frac{2 \times (4.2 \times 10^{3}) \times 50}{840}

= 500 s

= 8 min 20