This probllem is an example of how the total entropy of a sy

This probllem is an example of how the total entropy of a system increases simply when performing everyday tasks. There is a glass with 250 grams of room temperature (20 C) water in it. To this water is added 40 grams of 0C ice (this means the ice is just about to melt). The glass is then quickly plaved inside a well-insulated thermos, whihc means that no heat will be exchanged with the outside world (this makes the problem simpler to solve)>

a) How much heat will it take to melt all 40 grams of ice? latent heat of ice is 334 kJ/kg

b) Determine the final temperature of this mixture after all the ice has melted. Note that there are three heats involved: the heat it takes to lower the temperature of the water, the heat it heats to melt the ice, and the heat it takes to raise the temperature of the previously melted ice up to the final temperature. Specific heat of water is 4.184 kJ/kg

c) Calculate the total entropy change.Since there are three pocesses, two processes invove temperature change, while the ice melts, it remains at constant temperature (0C)

Solution

(a) heat absorbed is mL

here m is the mss of ice and th L is the latent heat

m=40 gram= 40*10^-3 kg

L=334kj/kg

so mL= 40*10^-3 *334=13.36 kj

(b) let T is the final temperaturre

heat absorbed by the ice to come to this temperature (i.e.T)

mL+ms(T-0) = .13.36 +40*10^-3*4.184*T..................(1)

heat released by the water to come to this temperature(T),

m₁s (20-T)= .........(2) here m₁ is the mass of water=0.250kg

0.250*4.184*20 - 0.250*4.184*T.....................(2)

equating (1) and (2)

13.36+0.167T=20.92 - 1.046T

1.213T=7.56

T= 6.23⁰C

(c) entropy = Q /T here Q represents heat and T is temperature.