00:01
For this problem on the topic of entropy, we are told that 0 .6 kilograms of water is initially ice at a temperature of minus 20 degrees celsius.
00:10
We want to know the entropy change for the sample if its temperatures increase to 40 degrees celsius.
00:17
Now, as the ice warms, the energy receives us heat when the temperature changes by dt is dq, which is its mass times the specific heat capacity for ice ci.
00:30
Times the temperature change d t now the initial temperature t i is given to be minus 20 degrees celsius which is 253 calvin and the final temperature t f is equal to 273 calvin and then it's change in entropy the delta s1 for the change in temperature is equal to the integral of dq by t, which is m -c -i times the integral from t -i to t -f of d -t by t, which is m -c -i times the natural log of d -t -by -t, which is m -c -i times the natural log of t f by t i and since we know these values we get this change in entropy to be 0 .6 kg times the specific heat capacity for ice which is 2 ,220 jule per kg kelvin times the natural log of 273 kelvin over 253 kelvin, which gives us the change in entropy to be 101 joules per kelvin going from minus 20 degrees celsius to 0 degrees celsius.
02:23
Now melting is an isothermal process and the energy leaving the ice as heat is mlf where lf is the latent heat of fusion for ice.
02:35
The change in entropy for this process, delta ace s2, is key...