W= Q_hot(1-(T_c/T_h)
that is the work done by the engine? is the Q_hot the wattage of the light bulb? and if you divide this by the time interval, 1 second, it wouldn't change the equation, right?
oh my gosh, I'm sorry! I'm not very good at physics!
ok, let's see... this is really frustrating me!
so if
so if (1-(T_c/T_h)=W/Q_hot, then Q_hot= W/(1-(T_c/T_h)) (that is the energy given off by the light bulb) so it's W+(W/(1-(T_c/T_h))? is that right? how do you find W?
ok... so the efficiency of a carnot engine is 1-(T_c/T_h), correct? so, the power, P, is W_out=P_out(delta T)/P_h (delta T)
so, wouldn't the equation be (P/(1-(T_c/T_h)))-P ??
i'm sorry, i still don't see what I'm doing wrong...
The inside of an ideal refrigerator is at a temperature , while the heating coils on the back of the refrigerator are at a temperature . Owing to a malfunctioning switch, the light bulb within the refrigerator remains on when the the door is closed. The power of the light bulb is ; assume that...