Recent content by bugonwndshld

ooh, i got it! the answer was -P(1-(T_h/T_c)), we needed to switch the c and h because we were dividing... yay! thanks so much for your help!

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...