What is the Entropy Change When Water Freezes to Ice?

[Another]

Homework Statement

10 kg water temperature 293 K change to ice ( 263 K) at constant pressure. Calculate the entropy change of system
Answer is $-13.8×10^3 J/°C$

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

My answer incorrect. Please help

 

[mjc123]

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.

 

[Another]

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]

I still agree with your answer. Textbooks can be wrong.

 

[Another]

I still agree with your answer. Textbooks can be wrong.

thank you very much