# How to find change in entropy for heat transfer between an object dropped in water

## 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)?

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 :)

Thank You!