## Related questions with answers

A car air-conditioning unit has a 0.5-kg aluminum storage cylinder that is sealed with a valve. It contains 2 L of refrigerant R134a at 500 kPa, and both are at room temperature, 20$^{\circ} \mathrm{C}$. It is now installed in a car sitting outside, where the whole system cools down to ambient temperature at −10$^{\circ} \mathrm{C}$. What is the irreversibility of this process?

Solution

VerifiedWe are given following data for air conditioner made of aluminum and contains R-134a:

$T_{1}=20\text{ C}=293\text{ K}$

$T_{2}=T_{amb}=-10\text{ C}=263\text{ K}$

$P=500\text{ kPa}$

$m_A=0.5\text{ kg}$

$V_R=2\text{ L}=0.002\text{ m}^3$

From Properties tables we can find Aluminum and R-134a specific heat:

$C_R=0.852\frac{\text{ kJ}}{\text{ kg K}}$

$C_A=0.90\frac{\text{ kJ}}{\text{ kg K}}$

$\rho_R=4.20\frac{\text{ kg}}{\text{ m}^3}$

Calculating R-134a mass:

$m_R=\rho_R\cdot V_R=4.20\cdot 0.002=0.0084\text{ kg}$

Heat transfer is equal to:

$\begin{align*} Q&=m_A\cdot C_A\cdot (T_2-T_1)+m_R\cdot C_R\cdot (T_2-T_1)\\\\ &=0.5\cdot 0.90\cdot (-10-20)+4.20\cdot 0.852\cdot (-10-20)\\\\ &=-13.7\text{ kJ} \end{align*}$

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