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Water of volume 2 liter in a container i...

Water of volume 2 liter in a container is heated with a coil of 1 kW at `27^(@)C`. The lid of the container is open and energy dissipates at the rate of ` 160 J s^(-1)` . In how much time, temperature will rise from ` 27^(@) C "to" 77^(@) C ` ?

A

8 min 20 s

B

6 min 2 s

C

2 min

D

14 min

Text Solution

Verified by Experts

The correct Answer is:
A
Rate at which energy is gained by water
= Rate at which energy supplied
- Rate at which energy is lost
` = 1000 j s^(-1) - 160 J s^(-1) = 840 J s^(-1) `
Heat required to raise the temperature of water from ` 27^(@) C "to " 77^(@) C " is ms " Delta T `
Hence, the required time
` t = ( ms Delta T )/( " Rate by which energy is gained by water ")`
` ( 2kg xxs 4.2 xx 10^(3) J kg^(-1@)C^(-1) xx (77^(@)C - 27^(@)C))/( 840 J s^(-1)`
` 500 s = 8 min 20 s `
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