Question:medium

Water ($C_{\text{p}} = 4.18\text{ kJ/kg}\cdot\text{K}$) at $80^\circ\text{C}$ enters a counter-flow heat exchanger with a mass flow rate of $0.5\text{ kg/s}$. Air ($C_{\text{p}} = 1\text{ kJ/kg}\cdot\text{K}$) enters at $30^\circ\text{C}$ with a mass flow rate of $2.09\text{ kg/s}$. If the effectiveness of the heat exchanger is 0.8, the LMTD ($^\circ\text{C}$) is

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For any balanced counter-flow heat exchanger ($C_{\text{h}} = C_{\text{c}}$), the temperature difference is uniform along the length.
The LMTD is equal to this constant temperature difference, which can be computed directly as $\Delta T = (1 - \epsilon)(T_{\text{h,in}} - T_{\text{c,in}})$.
Applying this formula here: $(1 - 0.8)(80 - 30) = 0.2 \times 50 = 10^\circ\text{C}$.