Saved Bookmarks
| 1. |
Ram ans Shyam purchased two electric tea kettles A and B of same size, same thickness, and same volume of 0.4L. They studied the specification of kettles as under Kettle A: specific heat capacity = 1680 Jkg^(-1)K^(-1) Mass = 200 g Cost = Rs. 400 Kettle B: Specific heat capacity= 2450 Jkg^(-1) K^(-1) Mass = 400 g Cost = Rs. 400 When kettle A is switched on with constant potential source, the tea begins to boil in 6 min. When kettle B is switched on with the same source separately, then tea begins to boil in 8 min. The efficiengy of kettle is defined as (Energy used for liquid heating)/(Total energy supplied) They made discussion on specification and efficiency of kettles and subsequently prepared a list of questions to draw the conclusions. Some of them are as under (assume specific heat of tea liquid as 4200J kg^(-1)K^(-1) and density 1000 kgm^(-3) If resistances of coil of kettles A and B are R_(A) and R_(B), respectively, then we can say |
|
Answer» `R_AgtR` `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 6.4(theta-theta_1)) xx 100 = 62.5%` `H_(A) = 0.2 xx 1680 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx [0.8 +4] (theta - theta_1)` `H_(B) = 0.4 xx 2450 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx (2.4 + 4)(theta-theta_1)` . `H_(A)/H_(B) = 4.8/6.4 = 3/4` `H_(A) = V^2/R_(A) xx 6` `H_(B) = V^2/R_(B) xx 8` `H_(A)/H_(B) = R_(B)/R_(A) xx 3/4 but H_(A)/H_(B) = 3/4 ` Then `R_(B) = R_(A)` |
|