# What is the Entropy Change When Water Freezes to Ice?

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In summary, the conversation discusses the calculation of entropy change for a system where 10 kg of water at 293 K changes to ice at 263 K at constant pressure. The correct answer is calculated to be -15968.4 J/°C, taking into account the specific heat capacities of water and ice, as well as the latent heat of fusion. It is noted that the textbook may have a typo in the answer provided.
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## Homework Statement

10 kg water temperature 293 K change to ice ( 263 K) at constant pressure. Calculate the entropy change of system

## Homework Equations

##c_p = 4180 J/kg-K## (water)
##c_p = 2090 J/kg-K## (ice)
##l_{water→ice} = 3.34×10^5 J/Kg##
##ΔS = \int \frac{dQ}{T} ##

## The Attempt at a Solution

##ΔS_1 = \frac{ml}{T}## ;Latent heat ##Q=ml##
##ΔS_1 = \frac{10×3.34×10^5}{273}=12234.4 J/K = 12234.4 J/°C##

##J/°C## equal to ##J/K##
http://www.endmemo.com/convert/specific heat capacity.php

I thank ##ΔS_1 ## should be negative. because water change to ice (lost heat)
So ## ΔS_1 = -12234.4 J/°c##

##ΔS_2= \int \frac{dQ}{T} = mc_p\int \frac{dT}{T}## integral 293 K to 273 K
##ΔS_2= mc_p \ln \frac{273}{293} = 10×4180×\ln \frac{273}{293} = -2955 J/°C##

##ΔS_3= mc_p\int \frac{dT}{T}## integral 273 K to 263 K
##ΔS_3= mc_p \ln \frac{263}{273} = 10×2090×\ln \frac{263}{273} = - 780 J/°C##

So ##ΔS_{sys} = ΔS_1+ΔS_2+ΔS_3 = -12234.4 -2955 - 780 = -15968.4J/°C##

Last edited:
I think you are correct. The answer of -13.8 x 105 is equal to - (10 x 4180 x 20) - (10 x 2090 x 10) - 3.34 x 105. So it looks as if they have calculated ΔH, but forgotten to multiply the latent heat by 10 kg.

mjc123 said:
I think you are correct. The answer of -13.8 x 105 is equal to - (10 x 4180 x 20) - (10 x 2090 x 10) - 3.34 x 105. So it looks as if they have calculated ΔH, but forgotten to multiply the latent heat by 10 kg.

I’m sorry. It was my fault. The answer of a question is ##-13.8×10^3 J/°C## I typed wrong. am so sorry

mjc123 said:
thank you very much

## What is the definition of freezing water?

Freezing water is the process in which liquid water changes into solid ice due to a decrease in temperature.

## At what temperature does water freeze?

Water freezes at a temperature of 0 degrees Celsius or 32 degrees Fahrenheit.

## What is the role of thermodynamics in freezing water?

Thermodynamics is the branch of science that deals with the study of energy and its transformations. In the case of freezing water, thermodynamics helps us understand the energy changes that occur during the phase transition from liquid to solid.

## Why does water expand when it freezes?

Water expands when it freezes because of its unique molecular structure. When water freezes, the molecules arrange themselves in a specific pattern, causing them to take up more space and expand.

## Can water freeze at temperatures above 0 degrees Celsius?

Yes, water can freeze at temperatures above 0 degrees Celsius, but only under certain conditions. For example, if the water is under high pressure, it can remain in liquid form at temperatures below its freezing point. This is known as supercooling. However, once the pressure is released, the water will freeze immediately.

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