Recent content by knc

Thanks for the help.

$$ Q_{net} = - W_{net} $$ $$ Q_a + Q_b + Q_c = -W_a - W_b - W_c $$ $$ \Delta E_{int,c} = \Delta E_{int,a} = 0 $$ $$ Q_a = -W_a \\ Q_c = - W_c $$ $$ -W_a + Q_b - W_c = -W_a - W_b - W_c $$ $$ Q_b = -W_a $$ Mathematically it makes sense. What physical implications does this hold?

So is it correct that the heat transfer in step B is equal to opposite of work done in step A? $$ Q_b = - W_a $$

So in step C, ##\Delta E_{int} = 0## $$ Q_c + W_c = 0 \\ Q_c = - W_c $$ Using integral to solve for W_c: $$ W_c = - \int_{V_2}^{4V_2} \frac{nRT}{V} dV \\ W_c = - nRT \ln{4} $$ How did you get positive work done? The piston is being raised, so isn't the gas doing work on its surroundings? Hence...

Does this mean the work done in step A is equal-but-opposite to the heat transfer in step B? Does this make sense?

## W_a + W_c = W_{net} \\ W_a = m L_f - W_c \\ W_c = -nRT \ln{4} \\ W_a = m L_f + nRT \ln{4} ##

So is the question misworded? By convention, I thought that a negative heat represented heat leaving a system whereas positive heat represented heat flowing into a system? Edit: Nevermind, it makes sense. The energy is transferring out of the gas so we denote the heat flow to be negative...

Could we apply this same model to solve for work done in step A, seeing as there is only work done in steps A and C?

Sorry for the late response... step C: ## \Delta E_{int} = 0 \\ W = -Q \\ \therefore Q = nRT \ln{4} ## This is our heat transfer in step C. Heat in step B: ## Q = mL_f - nRT \ln{4} ## Does this seem right?

In step C, since the piston is being raised slowly, we can say that the temperature is always in equilibrium, correct? So the work done on the gas in step C: W = - \int_{V_2}^{4V_2} \frac{nRT}{V}dV \\ W = - nRT \ln{4} How can we relate this to the heat transfer in step C? I understand your...

Of course I like this site better but when it's Sunday afternoon I cannot leave any stone unturned. :) We have determined in step A that there is no heat flow out of the system. But we still cannot use the integral to solve for work?

I see. Although the temperature is not constant, the process is done so quickly as to not transfer heat?

In step A the piston compressing the gas; it's going from its original volume (4V_2) to its compressed volume (V_2). Even then, are we assuming that the gas is compressed so quickly that the temperature remains constant? That doesn't seem right. Also, the hint for the problem in part B threw me...

Homework Statement Homework Equations PV = nRT \\ W = - \int_{V_i}^{V_f} P dV \Delta E_{int} = Q + W The Attempt at a Solution a)[/B] Since this is a cyclic process, the change in internal energy of the system is 0. \Delta E_{int} = 0 The process causes some ice to melt, meaning heat...

Thanks for the help.