Thermodynamics Numericals

JEE Advanced

1. Net work done by the gas during the cycle:

For the isothermal process AB, work done is $$W_{AB}=-P_AB(V_B-V_A)$$

For the adiabatic process BC, work done is $$W_{BC}=0$$ (As there is no volume change)

For the isothermal process CD, work done is $$W_{CD}=P_CD(V_D-V_C)$$

For the adiabatic process DA, work done is $$W_{DA}=0$$ (As there is no volume change)

Therefore, net work done during the cycle is: $$W_{net}=W_{AB}+W_{BC}+W_{CD}+W_{DA}=-P_AB(V_B-V_A)+P_CD(V_D-V_C)$$

Substituting the given values: $$=-(10\text{ atm})(20\text{ L}-10\text{ L})+(20\text{ atm})(10\text{ L}-20\text{ L})=-100\text{ atm L}+200\text{ atm L}=100\text{ atm L}$$

2. Work done by the gas during the process: $$W=P_e(V_e-V_i)$$

Substituting the given values: $$W=(10\text{ atm})(20\text{ L}-10\text{ L})=100\text{ atm L}$$

3. Final temperature of the mixture and amount of ice that melts: Let the final temperature of the mixture be $$T_f$$. Then, we have:

Heat gained by ice $$=m_{ice}L_f+m_{ice}C_{p,ice}(T_f-0°C)$$

Heat lost by water $$=m_{water}C_{p,water}(25°C-T_f)$$

where, $$m_{ice} = 100\text{ g}$$ $$m_{water}=100\text{ g}$$ $$L_f= 334 J/g$$ (latent heat of fusion of ice) $$C_{p,ice} = 2.09 J/g°C$$ (specific heat capacity of ice) $$C_{p,water} = 4.18 J/g°C$$ (specific heat capacity of water)

Setting the heat gained by ice equal to the heat lost by water, we get: $$m_{ice}L_f+m_{ice}C_{p,ice}(T_f-0°C)=m_{water}C_{p,water}(25°C-T_f)$$ $$(100 g)(334 J/g)+(100g)(2.09 J/g °C)(T_f)= (100g)(4.18 J/g °C)(25°C-T_f)$$ $$33400J+209T_f= 10450J-418T_f$$ $$623T_f=22950 J$$ $$T_f= \frac{22950 J}{623}=36.8°C$$

Amount of ice that melts:

$$m_{ice, melted}= m_{ice}- m_{ice, remaining}$$

$$m_{ice, remaining}=\frac{m_{ice}L_f}{C_{p,water}(T_f-0°C)}$$

$$m_{ice, remaining}= \frac{100g \times 334 J/g}{4.18J/g°C\times 36.8°C}=21.6 g$$

$$\therefore m_{ice, melted}= 100g - 21.6 g= 78.4 g$$

4. Maximum efficiency of the engine: $$\eta=1-\frac{T_L}{T_H}$$

Substituting the given values: $$\eta= 1-\frac{27°C+273}{127°C+273}=0.83$$

CBSE Board Exams

1. Work done by the gas:

$$W=P(V_f-V_i)$$

Substituting the given values: $$W=(10\text{ atm})(20\text{ L}-10\text{ L})=100\text{ atm L}$$

2. Final temperature of the mixture:

Let the final temperature of the mixture be $$T_f$$. Then, we have:

Heat gained by copper $$=m_{copper}C_{p,copper}(T_f-100°C)$$

Heat lost by water $$=m_{water}C_{p,water}(25°C-T_f)$$

where, $$m_{copper} = 50\text{ g}$$ $$m_{water}=100\text{ g}$$ $$C_{p,copper} = 0.39 J/g°C$$ (specific heat capacity of copper) $$C_{p,water} = 4.18 J/g°C$$ (specific heat capacity of water)

Setting the heat gained by copper equal to the heat lost by water, we get: $$m_{copper}C_{p,copper}(T_f-100°C)=m_{water}C_{p,water}(25°C-T_f)$$ $$(50 g)(0.39 J/g °C)(T_f-100°C)= (100g)(4.18 J/g °C)(25°C-T_f)$$ $$19.5T_f -1950J= 10450J-418T_f$$ $$437.5T_f=12400 J$$ $$T_f= \frac{12400 J}{437.5}=28.3 °C$$

3. Maximum coefficient of performance of the refrigerator:

$$COP_{ref}=\frac{T_L}{T_H-T_L}$$

Substituting the given values: $$COP_{ref}=\frac{5°C+273}{35°C+273-5°C+273}=5.6$$

4. Minimum coefficient of performance of the heat pump:

$$COP_{hp}=\frac{T_H}{T_H-T_L}$$

Substituting the given values: $$COP_{hp}=\frac{20°C+273}{-10°C+273-20°C+273}=2.8$$

Tips for Solving Thermodynamics Numericals

1. Clearly understand the thermodynamic process and the given conditions.
2. Identify the relevant formula or equation to be used.
3. Substitute the given values into the formula and calculate the required quantity.
4. Pay attention to the units of the given and calculated quantities.
5. Check the reasonableness of the obtained result.