- #1

Another

- 104

- 5

## 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

Last edited: