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Question 1 of 5
A rigid tank contains 2 kg of air initially at 300 K and 100 kPa. 100 kJ of heat is added to the air while the volume remains constant. Determine the final temperature of air. Assume air as an ideal gas with \( C_v = 0.718 \, kJ/kg\cdot K \).
Why: For closed system at constant volume, no work is done (\( W = 0 \)). First law: \( Q = \Delta U = m C_v \Delta T \). Substitute given values: \( 100 = 2 \times 0.718 \times (T_2 - 300) \). Solve for \( T_2 = 369.64 \, K \) or approximately 370 K.
Question 2 of 5
In a steady flow process through a device with one inlet and one outlet, 50 kJ/kg of heat is added and 30 kJ/kg of work is produced. The inlet enthalpy is 300 kJ/kg and inlet velocity is 10 m/s. Neglect potential energy changes and assume outlet velocity is 15 m/s. Find the outlet enthalpy.
Why: Steady flow energy equation: \( h_1 + q + \frac{V_1^2}{2} = h_2 + w + \frac{V_2^2}{2} \). Kinetic energy terms are small but included. Rearrange to solve for \( h_2 \).
Question 3 of 5
Which of the following statements is TRUE regarding the first law of thermodynamics?
A. It is applicable only to closed systems
B. For open systems, it accounts for mass flow energy
C. Heat and work are path functions
D. Internal energy is not a state function
A It is applicable only to closed systems
B For open systems, it accounts for mass flow energy
C Heat and work are path functions
D Internal energy is not a state function
Why: First law applies to both closed and open systems. For open systems, it includes enthalpy (\( h = u + pv \)) to account for flow work and energy carried by mass. Heat and work are path functions, but internal energy is a state function. Thus, B is correct.
Question 4 of 5
Derive the steady-state energy equation for an open system with one inlet and one outlet.
Why: Derivation starts from closed system first law and modifies for mass flow. Flow work \( pv \) combines with internal energy \( u \) to form enthalpy \( h = u + pv \). Steady-state assumption eliminates accumulation term.
Question 5 of 5
For the constant pressure process shown in the P-V diagram below, calculate the heat transfer and change in internal energy when 1 kg of ideal gas changes from state 1 to state 2.
Why: Constant pressure process: boundary work \( W = P(V_2 - V_1) \). First law closed system: \( Q = \Delta U + W \). Internal energy depends only on temperature change.