InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A heating coil of `2000 W` is immersed in water . How much time will it take in raising the temperature of `1 L` of water from `4^(@) C "to" 100^(@) C`? Only `80%` of the thermal energy produced is used in raising the temperature of water. |
|
Answer» Here, `P = 2000 W, t = ?` (in seconds) Volume of water = 1 litre = 1000 c.c. Mass of water, m = volume `xx` density `1000 xx 1 = 1000` gram Rise in temperature, ` = theta_(2) - theta_(1) = 100 - 4` `96^@C` We know sp. Heat of water, `s= 1 cal g^(-1) @C^(-1)` `:.` Heat taken by water `= ms (theta_(2) - theta_(1))` `= 1000 xx 1 xx 96 = 96000` cal. Energy spent in heating coil = pt `= 2000 xx t` Useful thermal energy produced `= 80 %` ` = 2000 xx t xx 80/100 = ( 2000 xx t xx 80)/( 100 xx 4.2)` cal As this thermal energy is taken by water, therefore, `( 2000 xx t xx 80)/(100 xx 4.2) = 96000` or `t (96000 xx 100 xx 4.2)/(2000 xx 80)` `= 252 seconds`. |
|