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Water of volume 2 litre in a container is heated with a coil of `1kW at 27^@C.` The lid of the container is open and energy dissipates at rate of `160J//s.` In how much time temperature will rise from `27^@C to 77^@C` Given specific heat of water is `[4.2kJ//kg`]A. `8 min 20 s`B. `6 min 2 s`C. `7 min`D. `14 min` |
Answer» Correct Answer - A Heat gained by the water = (Heat supplied by the coil) - (Heat dissipated to enviroment) `rArr mc Delta theta = P_("Coil") t - P_("Loss")t` `rArr 2 xx 4.2 xx 10^(3) xx (77-27) = 1000t = 160t` `rArr t = (4.2xx10^(5))/(840) = 500 sec = 8 min 20 sec`. |
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