# Calculating the change in entropy, did I calculate correctly?

• PhyIsOhSoHard
In summary, the change of entropy for the system of a 3.5 kg block of copper at 100 degrees Celsius (373 K) placed in 0.8 kg of water at 0 degrees Celsius (273 K), resulting in an equilibrium temperature of 30 degrees Celsius (303 K), is +65 J/K. This is calculated by finding the change of entropy for each component (copper and water) and adding them together. The calculation involves using the equations ΔS = Q/T and Q = mcΔT. It is important to keep extra significant digits in the intermediate calculations before rounding to the appropriate number of significant figures at the end.

## 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

$ΔS=\frac{Q}{T}$
$Q=mcΔT$

## The Attempt at a Solution

Change of entropy for the cobber block:
$ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=3.5kg\cdot 390\frac{J}{kg K}\left ( \ln\frac{303 K}{373 K} \right )=-284 J/K$

Change of entropy for the water:
$ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=0.8kg\cdot 4190\frac{J}{kg K}\left ( \ln\frac{303 K}{273 K} \right )=349 J/K$

Total change of entropy:
$ΔS=-284 J/K+349 J/K=+65 J/K$

Hello PhyIsOhSoHard,

PhyIsOhSoHard said:

## 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

$ΔS=\frac{Q}{T}$
$Q=mcΔT$

## The Attempt at a Solution

Change of entropy for the cobber block:
$ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=3.5kg\cdot 390\frac{J}{kg K}\left ( \ln\frac{303 K}{373 K} \right )=-284 J/K$

Change of entropy for the water:
$ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=0.8kg\cdot 4190\frac{J}{kg K}\left ( \ln\frac{303 K}{273 K} \right )=349 J/K$

Total change of entropy:
$ΔS=-284 J/K+349 J/K=+65 J/K$

Your approach looks correct to me.

However, I would be careful with rounding numbers too early. For example, some of the sources I see online for the specific heat capacity for copper are around 386 J/(Kg K) rather than 390.

As a rule of thumb, try to keep a couple of extra significant digits around in the intermediate calculations, then round to the appropriate number of significant figures at the very end.

## 1. How do I calculate the change in entropy?

The change in entropy can be calculated by subtracting the final entropy value from the initial entropy value. This can be represented as ΔS = Sf - Si, where ΔS is the change in entropy, Sf is the final entropy value, and Si is the initial entropy value.

## 2. What units are used to measure entropy?

Entropy is typically measured in joules per Kelvin (J/K) in the SI system. However, it can also be measured in other units such as calories per Kelvin (cal/K) or kilojoules per Kelvin (kJ/K).

## 3. Can entropy have a negative value?

Yes, entropy can have a negative value. This occurs when there is a decrease in disorder or randomness in a system. An example of this is when a gas is compressed, causing the molecules to become more ordered and decreasing the entropy of the system.

## 4. Is there a specific formula to calculate entropy?

Yes, the formula for calculating entropy is ΔS = q/T, where ΔS is the change in entropy, q is the heat transferred in a reversible process, and T is the temperature in Kelvin. This formula is known as the Clausius-Clapeyron equation.

## 5. What are some common mistakes when calculating entropy?

One common mistake when calculating entropy is not converting the temperature to Kelvin. Entropy is dependent on temperature, which must be in Kelvin for accurate calculations. Another mistake is not taking into account the reversible nature of the process, which can lead to incorrect calculations. Additionally, using the wrong units or using different units for different values in the equation can also result in errors.