An 8.0 g ice cube at is put into a Thermos flask containing of water at . 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 .
Mass of ice is
Initial Temperature of ice
Specific heat of ice
Mass of water
Initial Temperature of water
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,
The heat transferred by the body,
For an equilibrium state, the heat lost by of water is absorbed by ice which melts and reaches temperature
For of water, the specific heat is given by
When the equilibrium is reached, then
During the equilibrium process, warm water at cools to a final temperature , ice at temperature warms to , and this ice first melts, and then melted ice (water) warms.
The specific heat of water is given by the formula of equation (2) as given:
is heat fusion of water, and its value is given by
Substituting all values in the above expression, we get that the final temperature can be obtained as,
Entropy change of two blocks coming to equilibrium is given by using equation (i) can be given as:
For the phase change experienced by the ice the total entropy change is given as:
Hence, the value of the entropy change of the system is
Figure 20-27 shows a reversible cycle through which 1.00 mol of a monatomic ideal gas is taken. Assume that , and . (a) Calculate the work done during the cycle, (b) Calculate the energy added as heat during stroke abc, and (c) Calculate the efficiency of the cycle. (d) What is the efficiency of a Carnot engine operating between the highest and lowest temperatures that occur in the cycle? (e) Is this greater than or less than the efficiency calculated in (c)?
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