Find the net entropy change when a 50-g ice cube initially at 0^∘ C melts in (a) a 10^∘ C room a

Find the net entropy change when a 50-g ice cube initially...

Key Concepts

Entropy

Entropy is a measure of the disorder or randomness in a system and it quantifies how much energy is unavailable for doing work. In thermodynamic processes, entropy changes are calculated by the reversible heat transfer divided by the temperature at which the transfer occurs, highlighting how energy disperses in the system.

Latent Heat of Fusion

The latent heat of fusion is the amount of energy required to change a substance from a solid phase to a liquid phase at a constant temperature. For melting ice, this energy input does not change the temperature but increases the entropy of the ice as it transforms into water.

Heat Transfer Between System and Surroundings

In any thermodynamic process, heat transfer between the system (in this case, the melting ice) and its surroundings plays a crucial role in determining the net entropy change. The entropy change of the surroundings is determined by the amount of heat exchanged divided by the temperature of the reservoir, emphasizing that higher temperatures lead to smaller entropy penalties for heat loss.

Overall Net Entropy Change

The overall net entropy change for a process accounts for both the increase in entropy due to the phase change within the system and the change in entropy of the surroundings due to heat transfer. This concept, governed by the second law of thermodynamics, ensures that the total entropy change for any spontaneous process is positive.

Transcript

00:01 So this question belongs to the loss of thermodynamics in which we have to calculate the net entropy change delta as when 50 gram ice, 50 gram ice initially at 0 degree centigrade t initial it is equal to 0 degree centigrade melts and for the part a at the 10 degree centigrade temperature.

00:19 So first of all calculating the entropy change for the melting, so it will be equal to melting it will be equals to delta q divided by temperature t.

00:29 So, delta q for the melting, it will be m multiplied by l divided by temperature t.

00:33 So m is equal to 50 gram, that is 0 .050 kilogram and latent heat of vaporization, it is triple three multiplied by 10 to the power 3 jule per kilogram divided by temperature is 0 degree centigrade, that is 273 calvite.

00:50 So from here we get delta s for the melting, it is equals to 60 .98 jule per kelvin.

00:58 Now, for the part a, the temperature is at t, it is equal to 10 degrees centigrade that is equal to 283 kelvin.

01:08 So the heat involved in this process will be, it is equal to, so we can write that entropy for the, this process, delta s, it is equals to q divided by temperature t.

01:23 So q, it is this value, okay? so this value and this value will be equals to and this value is equal to 1 .665 multiplied by 10 to the power 4 jule divided by temperature which is 283 kelvin and this heat is in the negative because the heat is being absorbed by the air.

01:48 So from here we get delta s it is equals to minus 58 .83 jule per kelvin.

01:54 So, net change in entropy delta s, net, it will be equals to delta s melting plus delta s this value...

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