The temperature inside a refrigerator is t_(2)^(@)C and the room temperature is t_(1).^(@)C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

The temperature inside a refrigerator is t_(2)^(@)C and the room tempe

1.	The temperature inside a refrigerator is t_(2)^(@)C and the room temperature is t_(1).^(@)C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
Answer» `(t_(1))/(t_(1) - t_(2))` `(t_(1) + 273)/(t_(1) - t_(2))` `(t_(2) + 273)/(t_(1) - t_(2))` `(t_(1) + t_(2))/(t_(1) + 273)` Solution :Temperature INSIDE refrigerator `= t_(2).^(@)C` Room temperature `= t_(1)^(@)C` For refrigerator, `("Heat given to high temperature "(Q_(1)))/("Heat taken from lower temperature "(Q_(2))) = (T_(1))/(T_(2))` `(Q_(1))/(Q_(2)) = (t_(1) + 273)/(t_(2) + 273)` `rArr (Q_(1))/(Q_(1) - W) = (t_(1) + 273)/(t_(2) + 273)` or `1 - (W)/(Q_(1)) = (t_(2) + 273)/(t_(1) + 273)` or `(W)/(Q_(1)) = (t_(1) - t_(2))/(t_(1) + 273)` The amount of heat delivered to the room for each JOULE electric ENERGY (W= 1J) `Q_(1) = (t_(1) + 273)/(t_(1) - t_(2))`