InterviewSolution
Saved Bookmarks
| 1. |
An electrically heating coil was placed in a calorimeter containing `360g` of water at `10^(@)C`. The coil consumes energy at the rate of `70W`. The water equivalent of calorimeter and coil is `40g` . Find temperature of the water after `10min`. Given specific heat of water = `1 cal//g.^(@)C, 1 cal = 4.2J`. |
|
Answer» The energy consumed by the coil in `10 min` `W =Pt = 70xx(10xx60) = 4.2xx10^(4)J` ..(i) Let final temperature of water = `10^(@)C` (given) Heat absorbed by (water+calorimeter + coil) `Delta Q = msDelta theta = (360+40)xx 11 xx(theta_(2)-10)cal` `=40(theta_(2)-10)xx4.2J` ...(ii) `Delta Q = W` `400(theta_(2)-10)xx4.2 = 4.2xx10^(4)` `theta_(2)-10=25` `theta_(2) = 35^(@)C`. |
|