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Fundamentals Of Physics
Found in: Page 605

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Short Answer

An 8.0 g ice cube at -10°C is put into a Thermos flask containing 100 cm3of water at 20°C. By how much has the entropy of the cube – water system changed when equilibrium is reached? The specific heat of ice is 2220J/kg.K

Entropy change of the cube-water system when equilibrium is reached is 0.64J/°C.

See the step by step solution

Step by Step Solution

Step 1: The given data

Mass of ice ismice=8gm10-3kg1 gm=0.008 kg

Initial Temperature of iceTinitialice=-10°C=263.15 kg

Specific heat of icecice=2220J/lg.K=2220J/kg.°C

Mass of watermwater=100 gm10-3 kg1 gm=0.1 kg

Initial Temperature of water Tinitialwater=20°C=293.15 K

Step 2: Understanding the concept of entropy change

We use the concept of entropy change of the two blocks coming to thermal equilibrium to calculate entropy. We use the formula of specific heat of the water to calculate the final temperature of the water.


The entropy change of the gas,

Stemp change=mc InTfinalTinitial (1)


The heat transferred by the body,

Q=cmT (2)

Step 3: Calculation of the entropy change of the cube-water system

0.64J/°CFor an equilibrium state, the heat lost by 100 cm3mwater of water is absorbed by ice mice which melts and reaches temperature Tf>0

For mwater=0.1 kg of water, the specific heat is given by

cwater=4.190J/g.°C =4190J/kg.°C

When the equilibrium is reached, then Q=0

During the equilibrium process, warm water at 20°C cools to a final temperature Tf, ice at temperature -10°C warms to 0°C, and this ice first melts, and then melted ice (water) warms.

Qwarm water cools+Qice warms to 0°C+Qice melts+Qmelted ice warms=0

The specific heat of water is given by the formula of equation (2) as given:


LF is heat fusion of water, and its value is given by LF=333×103Jkg

Substituting all values in the above expression, we get that the final temperature can be obtained as,

4190J/kg.°C×0.1 kgTf-4190J/kg.°C×0.1kg×20°C+2220 J/kg.°C×0.008kg×10°C+333×103 J/kg×0.008 kg+4190 J/kg.°C×0.008kg×Tf=0

419 J/°CTf-8380J+177.6 J+2841.6 J+33.52 J/°CTf=0452.52 J/°CTf=5538.4 JTf=12.24°CTf=285.39 K

Entropy change of two blocks coming to equilibrium is given by using equation (i) can be given as:

Stemp change=mcInT2T1Smelt=LFmT0

For the phase change experienced by the ice (withT0=273.15K) the total entropy change is given as:

Ssystem=mwatercwaterIn285.39K293.15K+miceciceIn273.15K263.15K +micecwaterIn285.39K273.15K+LFmice273.15K =0.1 kg4190J/kg.°CIn285.39K293.15K+ 0.008 kg2220J/kg.°CIn285.39K273.15K+2664J273.15K

Ssystem=-11.24+0.66+1.47+9.75J/°C =0.64J/°C

Hence, the value of the entropy change of the system is 0.64J/°C

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