Light lubricating oil (Cp = 1087 J/kg * K) is cooled by allowing it to exchange energy with water in a small heat exchanger. The oil enters and leaves the heat exchanger at 380 and 355 K, respectively, and flows at a rate of 0.6 kg/s. Water at 298 K is available in sufficient quantity to allow 0.301 kg/s to be used for cooling purposes. Determine the required heat-transfer area for counter flow operations. The overall heat-transfer coefficient may be taken as 200 W/m/ K. U / Answer heat transfer area in m^2 (meter squared)