Solving for Entropy Change: Copper & Water

[MeliMo]

Homework Statement

A 500 g copper block at 80 C is dropped into 1.0 kg of water at 10 C. Find the entropy change of the system.

Homework Equations

dS= integral of dQ/T

The Attempt at a Solution

I found that the final equilibrium temperature is 290 K using the equation
[(m1*c1*dT1) + (m2*c2*dT2)]/ [(m1*c1) + (m2*c2)]
I know that this is correct, but I'm not sure what to do from here. I know that this is Tf, but what do I use as Ti for the integral, the temperature of the block or the water? And which equation do I integrate to find the entropy in this case (What is the equation for Q)?

 

[nickjer]

Well, not sure why you wrote dT1 and dT2 in the equation. But you are correct if you meant:
$$T_f = \left(m_1 c_1 T_1 + m_2 c_2 T_2\right)/\left(m_1 c_1 + m_2 c_2\right)$$

You also know that $$dQ = m\cdot c\cdot dT$$ but forgot to mention it in your list of equations. Knowing this and that
$$\Delta S = \int_{T_i}^{T_f}dQ/T$$
you can solve for the change in entropy, $$\Delta S$$, for both systems. The total change in entropy will just be the sum of the two values.

P.S.: Be careful with units. And be sure the total change in entropy is greater than or equal to zero, we don't want to be breaking any laws here :)

 

[MeliMo]

Thank You!