An electrically heated coil is immersed in a calorimeter containing 360 g of water at `10^@C`. The coil consumes energy at the rate of 90 W. The water equivalent of calorimeter and coil is 40 g. The temperature of water after 10 min is

Options

A. `4.214^@C`
B. `42.14^@C`
C. `30^@C`
D. none of these

Answer

B

Explanation

Energy supplied by the heater to the system in 10 min
`Q_1=Pxxt90 J//s xx10xx60s`
`=54000J=(54000)/(4.2)cal=12857cal`
Now if `theta` is the final temeprature of the system, energy absorbed by it to change its temperature from `10^@C` to `theta^@C` is
`Q_2=(msDeltaT)_("water")+(msDeltaT)_("coil + calorimeter")`
`=360xx1xx(theta-10)+40(theta-10)`
`=400(theta-10)`
According to problem, `Q_1=Q_2`
So `12857=400(theta-10)` or `theta=42.14^@C`