Water of volume 2 litre 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. In how much time, temperature will rise from 27^(@)C" to "77^(@)C ? [Given specific heat of water is 4 cdot 2 kJ/kg]:

Water of volume 2 litre in a container is heated with a coil of 1 kW a

1.	Water of volume 2 litre 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. In how much time, temperature will rise from 27^(@)C" to "77^(@)C ? [Given specific heat of water is 4 cdot 2 kJ/kg]:
Answer» 8 min 20 s 7 min 6 min 2 s 14 min Solution :Heat produced by HEATER = Heat gained by water + heat lost DUE to radiation. `1000 t =mc DELTA T+160 t` `1000 t=2xx 4 cdot 2xx10^(3)xx(77-27)+160 t` `rArr t=500 s=8` min 20 sec. `therefore` CORRECT choice is (a).

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Water of volume 2 litre 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. In how much time, temperature will rise from 27^(@)C" to "77^(@)C ? [Given specific heat of water is 4 cdot 2 kJ/kg]:

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