Question

Air in a closed system undergoes a thermodynamic process from 300 K to 400 K. The specific heat at constant volume varies linearly with temperature: \[ c_v = (0.7 + 0.27\times10^{-3} T)\ \text{kJ/(kgK)} \] Change in specific internal energy is _________ kJ/kg (round off to 2 decimal places).

Hint:

When $c_v$ varies with temperature, integrate directly: $\Delta u = \int c_v(T)\, dT$.

Answer 1

Correct Answer: 79

Internal energy change: \[ \Delta u = \int_{T_1}^{T_2} c_v\, dT \] \[ \Delta u = \int_{300}^{400} \left(0.7 + 0.27\times10^{-3} T\right)\, dT \] Integrate: \[ \Delta u = 0.7(T)\Big|_{300}^{400} + 0.27\times10^{-3} \frac{T^{2}}{2}\Big|_{300}^{400} \] \[ \Delta u = 0.7(100) + 0.00027 \cdot \frac{400^2 - 300^2}{2} \] First part: \[ 0.7(100) = 70 \] Second part: \[ 400^2 - 300^2 = 160000 - 90000 = 70000 \] \[ 0.00027 \cdot \frac{70000}{2} = 0.00027 \cdot 35000 = 9.45 \] Total: \[ \Delta u = 70 + 9.45 = 79.45~\text{kJ/kg} \] \[ \boxed{79.45~\text{kJ/kg}} \]