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A 10 W electric heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of water and container rises by 3^(@)K in 15 min. The container is then emptied, dired and filled with 2kg of oil. The same heater now raises the temperature of container oil system by 2K in 20 min. Assume there is no heat loss in the process and the specific heat of water is 4200Jkgi^(-1)K%^(-1), the specific heat of oil in the same limit is equal to |
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Answer» <P>`1.50 xx 10^(3)` 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)` |
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