Saved Bookmarks
| 1. |
The inside and outside temperature of a refrigerator are 273 K and 303 K respectively. Assuming that refrigerator cycle is reversible for every joule of work done, the heat delivered to the surroundings will be |
|
Answer» Solution :As `beta = (Q_(2))/(W) = (T_(2))/(T_(1) - T_(2))` `:. Q_(2) = W(T_(2))/(T_(1) - T_(2)) = 1 xx (273)/(303 - 273) = (273)/(30) ~= 9J` Heat delivered to the SURROUNDINGS, `Q_(1) = Q_(2) + W = 9 + 1 = 10 J` |
|