- #1

PhyIsOhSoHard

- 158

- 0

## Homework Statement

A 3.5 kg block of cobber at 100 degrees celsius (373 K) is put in 0.8 kg water at 0 degrees celsius (273 K).

The equilibrium temperature is 30 degrees celsius (303 K).

Calculate the change of entropy for the system of cobber and water.

## Homework Equations

[itex]ΔS=\frac{Q}{T}[/itex]

[itex]Q=mcΔT[/itex]

## The Attempt at a Solution

Change of entropy for the cobber block:

[itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=3.5kg\cdot 390\frac{J}{kg K}\left ( \ln\frac{303 K}{373 K} \right )=-284 J/K[/itex]

Change of entropy for the water:

[itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=0.8kg\cdot 4190\frac{J}{kg K}\left ( \ln\frac{303 K}{273 K} \right )=349 J/K[/itex]

Total change of entropy:

[itex]ΔS=-284 J/K+349 J/K=+65 J/K[/itex]