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This is the second of our many quizzes on Entropy. Click on this to access more such competitive quizzes on Thermal Physics.

Entropy Quiz 2

physics

 If the enthalpy change of a reaction is $134$ kJ/mol and the temperature is $280$ K, find the entropy change that occurs during this reaction? 

 The relation between entropy and enthalpy is given by $\Delta S=\dfrac{\Delta H}{T}$.Putting the values of enthalpy and temperature in the above equation we calculate the value of entropy. 

 Entropy is the measure of disorder of the system and enthalpy is the total heat content of the system. $ \\ $ The relation between entropy and enthalpy is given by, $ \\ \Delta S=\dfrac{\Delta H}{T} \\ $ According to the question, $ \\  \Delta H=134\,kJ/mol\,\text{and}\, T=280\,K\\ $ Putting these values in the above equation, we get, $ \\  \Delta S=\dfrac{134000}{280}\\  \therefore\Delta S =478.6\,J$ 

To produce water at a temperature ${{60}^{0}}$C by using $x$ grams of ice and $40$ grams of steam at ${{100}^{0}}$C, find the value of $x$. The latent heat of fusion is $333\times {{10}^{3}}$ J/kg, specific heat capacity of water is $4190$ J/kgK and latent heat of vaporization is $2256\times {{10}^{3}}$ J/kg. 

 The heat change when water at a temperature ${{60}^{0}}$ C is made by mixing $x$ grams of ice and $40$ grams of steam at ${{100}^{0}}$ C should be equal, so put the values and equate heat change during melting of ice and cooling of steam to find the value of only unknown i.e.$x$, the grams of ice. 

 The formula for heat change of the system when ice changes into water is given by, $ \\ {{Q}_{i}}={{m}_{i}}{{L}_{f}} \\ $ And formula for heat change of the system when water is heated is given by, $ \\  Q=mc\Delta T\\ $ According to the question, water at a temperature ${{60}^{0}}C$ is made by mixing x grams of ice and 40 grams of steam at ${{100}^{0}}C.$ $ \\ $ Then, heat change in both the cases should be equal ${{Q}_{i}}={{Q}_{s}} \\  \Rightarrow {{m}_{i}}{{L}_{f}}+{{m}_{i}}{{c}_{w}}\Delta T={{m}_{s}}{{L}_{v}}+{{m}_{s}}{{c}_{w}} \Delta T….(i)\\ $ Where ${{m}_{i}}=xg $ is the mass of ice ${{L}_{f}}=333\times {{10}^{3}}\,J/kg$ is the latent heat fusion, ${{c}_{w}}=4190\,J/kgK,\,{{m}_{s}}=40g,{{L}_{v}}=2256\times {{10}^{3}}\,J/kg\\ $ And $\Delta T$ is the temperature difference. $\\ $ Putting these values in equation $(i)$ and solving, we get the mass of ice as, $ \\ \therefore x=165.89\,g$ 

If $155.03$ kJ heat is needed to increase the temperature of water at ${{299}^{0}}$ K to ${{373}^{0}}$ K, then what is the mass of water? $(\text{The specific heat of water is 4190 J})$

 Heat needed for the reversible isothermal process is given by $Q=mc\Delta T$.Putting the values given in the question calculate the mass of water. 

$  \text{Heat needed for reversible isothermal process is given by,}\\  Q=mc\Delta T….(i)\\  \text{According to the question,}\\  \text{c is}\,4190\,J,Q=155.03\,kJ\,\text{and}\,\Delta T=373-299={{74}^{0}}K\\  \text{Putting the above values in equation (i), we get,}\\  \therefore m=0.5\,kg=500\,g$ 

A glass of coca-cola has $200$ g of cola and $50$ g of ice at ${{0}^{0}}$C. What will be the entropy change when all the ice has melted? $(\text{Latent heat of fusion is}\, 333\times {{10}^{3}}\,J/kg)$ 

 The formula for entropy change of the system when ice changes into water are given by $\Delta S={{m}_{i}}{{L}_{f}}$. Putting the values of the mass of ice and latent heat fusion, we get entropy change of the system. 

The formula for entropy change of the system when ice changes into water is given by, $ \\  \Delta S={{m}_{i}}{{L}_{f}}….(i)\\ $ According to the question, $ \\  {{m}_{i}}=50g=50\times {{10}^{-3}}kg\,\text{and}\,{{L}_{f}}=333\times {{10}^{3}}\,J/kg.\\ $ Putting the above values in equation (i), we get, $ \\ \therefore\Delta S=16.65\,kJ$  

What is the molar heat capacity of liquid bromine, if there is an entropy change of $18.77$ kJ when $5$ g of bromine at $293$ K is heated to $343$ K. 

 Entropy change is given by $\Delta S=mc\Delta T$.Putting the values of the mass of bromine, entropy change and $\Delta T$, we get the molar heat capacity of liquid bromine. 

$\text{Entropy change is given by,}\\  \Delta S=mc\Delta T….(i)\\  \text{According to the question,}\\\Delta S=18.77\,kJ,m=5\,g=5\times {{10}^{-3}}kg\,\Delta T=343-293={{50}^{0}}K\\  \text{Putting the above values in equation (i),} \\ \text{We get the molar heat capacity of liquid bromine as,}\\\therefore c=75.1\,kJ/mol\cdot K$   