# Work Check: Heat transfer between reservoir and small system

• WWCY

## Homework Statement

Could someone look through my work? The parts where I wrote (??) are steps I am especially unsure about.

A large reservoir at temperature ##T_r## is placed in thermal contact with a small system at temperature ##T##. They end up at temperature ##T_r##. Given that ##C## is the heat capacity of the system, find the total change in entropy in the universe.

## The Attempt at a Solution

I started by looking for the entropy of the system. Since the volumes don't change, I can write
$$dU_{s} = TdS_{s}$$
Since ##U_{s}## is a state function that depends only on the initial and final states of the system, I take some reversible pathway (??), which means
$$dU_{s} = TdS_{s} = \delta Q_{rev}$$
$$dS_{s} = \frac{1}{T} C dT$$
Integrating gives me $$\Delta S_{sys} = C \ln{\frac{T_r}{T}}$$
Then for the entropy of the reservoir: Conservation of energy, and the fact that the reservoir doesn't change its volume allows me to write (again, using some reversible path [?])
$$dU_{r} = T_{r}dS_{r} = -dU_{s} = -CdT$$
$$dS_{r} = -\frac{1}{T_{r}}CdT$$
Integrating gives me ##\Delta S_{r} = \frac{C(T - T_r)}{T_r}##

The total change is then the sum of these two changes in entropy.

## Homework Statement

Could someone look through my work? The parts where I wrote (??) are steps I am especially unsure about.

A large reservoir at temperature ##T_r## is placed in thermal contact with a small system at temperature ##T##. They end up at temperature ##T_r##. Given that ##C## is the heat capacity of the system, find the total change in entropy in the universe.

## The Attempt at a Solution

I started by looking for the entropy of the system. Since the volumes don't change, I can write
$$dU_{s} = TdS_{s}$$
Since ##U_{s}## is a state function that depends only on the initial and final states of the system, I take some reversible pathway (??), which means
$$dU_{s} = TdS_{s} = \delta Q_{rev}$$
$$dS_{s} = \frac{1}{T} C dT$$
Integrating gives me $$\Delta S_{sys} = C \ln{\frac{T_r}{T}}$$
Then for the entropy of the reservoir: Conservation of energy, and the fact that the reservoir doesn't change its volume allows me to write (again, using some reversible path [?])
$$dU_{r} = T_{r}dS_{r} = -dU_{s} = -CdT$$
$$dS_{r} = -\frac{1}{T_{r}}CdT$$
Integrating gives me ##\Delta S_{r} = \frac{C(T - T_r)}{T_r}##

The total change is then the sum of these two changes in entropy.
Looks good. Nice job.