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`.

Explanation

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`.