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Problem 2: A 100 g ice cube, initially at 0°C, is placed on a table in a room at 20°C. The ice cube melts, and the water eventually comes into thermal equilibrium with the room. Assume that the temperature of the room doesn't change during this process. (a) How much heat is absorbed by the ice cube in order to melt it? (The "latent heat" of ice is 333 J/g, that is, 333 J are required to melt every gram of ice.) (b) What is the entropy change of the ice as it melts? (c) How much heat is absorbed by the water as it comes into thermal equilibrium with the room? (The specific heat of water is 4.2 J/g/°C) (d) What is the entropy change of the water as it comes to thermal equilibrium? (e) What is the entropy change of the room as it melts the ice then heats the water? (f) What is the total change in entropy of the universe during this process? Is it in accordance with the second law? Explain:

          Problem 2: A 100 g ice cube, initially at 0°C, is placed on a table in a room at 20°C. The ice cube melts, and the water eventually comes into thermal equilibrium with the room. Assume that the temperature of the room doesn't change during this process.

(a) How much heat is absorbed by the ice cube in order to melt it? (The "latent heat" of ice is 333 J/g, that is, 333 J are required to melt every gram of ice.)

(b) What is the entropy change of the ice as it melts?

(c) How much heat is absorbed by the water as it comes into thermal equilibrium with the room? (The specific heat of water is 4.2 J/g/°C)

(d) What is the entropy change of the water as it comes to thermal equilibrium?

(e) What is the entropy change of the room as it melts the ice then heats the water?

(f) What is the total change in entropy of the universe during this process? Is it in accordance with the second law? Explain:

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Added by Michael G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Neelesh Sharma

03/20/2022

Step 1

Given: Mass of ice cube, m = 100 g = 0.1 kg Latent heat of ice, L = 333 J/g The heat absorbed by the ice cube to melt it is given by: Q = m * L Q = 0.1 kg * 333 J/g Q = 33.3 kJ Therefore, the heat absorbed by the ice cube to melt it is \(\mathbf{33.3 kJ}\). Show more…

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Problem 2: A 100 g ice cube, initially at 0°C, is placed on a table in a room at 20°C. The ice cube melts, and the water eventually comes into thermal equilibrium with the room. Assume that the temperature of the room doesn't change during this process. (a) How much heat is absorbed by the ice cube in order to melt it? (The "latent heat" of ice is 333 J/g, that is, 333 J are required to melt every gram of ice.) (b) What is the entropy change of the ice as it melts? (c) How much heat is absorbed by the water as it comes into thermal equilibrium with the room? (The specific heat of water is 4.2 J/g/°C) (d) What is the entropy change of the water as it comes to thermal equilibrium? (e) What is the entropy change of the room as it melts the ice then heats the water? (f) What is the total change in entropy of the universe during this process? Is it in accordance with the second law? Explain:

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Transcript

00:01 Hello friends, so this is a question related to thermal equilibrium.

00:05 Now we have, in this question we have a mass of 100 grams of ice cube which is initially at 0 degrees celsius and is placed on the table, a room and the room temperature is 20 degrees celsius.

00:23 The ice cube melts and the water eventually comes thermal equilibrium with the room.

00:29 Assume that the temperature of the room does change during the process.

00:32 So the temperature remains constant at 20 and this one here it changed from zero.

00:38 So how much heat is absorbed by the tube in order to melt? the latent heat of ice is given as july june per gram.

00:53 So now we can use this formula directly.

00:58 So the total heat absorbed is equal to m into l which is the latent heat given at 3.

01:09 The mass is given as 100 multiplied by the latent heat.

01:14 That is 3333.

01:16 So it will be 33 .3 kilojoules.

01:21 So the first answer will be 33 .3 kilojoules is absorbed by ice for melting.

01:28 Next is what is the entropy change of the ice as it melts? so the change in entropy is given by dels.

01:37 Change in entropy is equal to change in field.

01:45 So change in heat divided by t temperature at least the change of course.

01:52 So assuming that this is here, 333 0 divided by the time takes, sorry, temperature at 2 .73, directly taking the 0 to celsius temperature temperature.

02:09 So at this, the change in entropy will be 121 .98 jule per kelvin...

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