Correct Answer - B
b) As we know, work done by an electric heater,
i.e., `m_(1)c_(1)Deltat+m_(2)c_(2)Deltat`= work done
So, `m_(1)c_(1)Deltat+m_(2)c_(2)Deltat=P_(1)t_(1)`
where, `m_(1) = 0.5` kg, specific heat `c_(1)` = 4200 J`kg^(-1)K^(-1)`,
`Deltat = Deltat_(1) = Deltat_(2)=3K`
`P_(1)=P_(2)`=10W
`t_(1)` = 15 xx60=900 s
`c_(2)`= specific heat capacity of container.
So, `0.5 xx4200xx(3-0) + m_(2)c_(2)xx(3-0) = 10 xx15 xx60`
`2100 xx3 +m_(2)c_(2)xx3=9000`
`m_(2)c_(2)=(900-6300)/3 = 900`
Similarly, in the cae of oil,
`m_(1)c_(0)Deltat + m_(2)c_(2)Deltat=P_(2)t_(2)`
Where, `c_(0)` = specific heat capacity of oil
`P_(1)=P_(2)`=10W
`2 xxc_(0)xx2+900 xx2=10 xx20 xx60`
`4c_(0)` + 1800 = 12000
`c_(0)` = 2550 = `2.55 xx 10^(3)`J`kg^(-1)K^(-1)`