Question

Two spaces are maintained at temperature difference of 200 K by a refrigerating unit. If the initial temperature of spaces are ${30}^{\circ }$C and specific heat 2 kJ/kg-K. What is the minimum work required to achieve the required temperature difference?

• A. 32 kJ/kg
• B. 64 kJ/kg
• C. 128 kJ/kg
• D. 16 kJ/kg

Answer 1

The correct option is B 64 kJ/kg





$T_0={30}^{\circ }C=303K$
$T_1-T_2=200K$
$Q_1=Q_2+W$ (First law of thermodynamics)
for minimum work process should be reversible.
$(\Delta S{)}_1+(\Delta S_2)=0$
$\Rightarrow$ CIn$\left(\frac{T_1}{T_0}\right)+CIn\left(\frac{T_2}{T_0}\right)=0$
$\Rightarrow$ $T_0^2=T_1{T}_2=(200+T_2)T_2$
$\Rightarrow$ $T_2^2+200T_2-{303}^2=0$
$T_2=219K,-419K$
Negative value is rejected because absolute temperature can not be negative.
$T_2=219K$
$T_1=419K$$W=Q_1-Q_2=C(T_1-T_0)-C(T_0-T_2)$
$W=2(T_1+T_2-2T_0)$
= 2(419 + 219 - 2 $\times$ 303) = 64 kJ/kg