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Water of volume 2 liter in a container is heated with a coil of 1 kWat 27^(@)C.The lid of the container is open and energydissipates at therate of160 J s^(-1) . In howmuch time,temperaturewill rise from27^(@) C "to"77^(@) C? |
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Answer» 8 MIN 20 s = Rate at whichenergysupplied - Rateat which energyis lost ` = 1000 j s^(-1)- 160 J s^(-1)= 840 J s^(-1) ` Heat REQUIREDTO raisethe temperatureof water from ` 27^(@) C"to "77^(@) C " is ms " Delta T ` Hence, the required time ` t = ( ms Delta T )/( " Rate by which energy is gainedby 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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