# Thermodynamics: freezing water

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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##

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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 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