4. Determine the power of an electric heater needed to heat oil to t=65^circ C in 1 h in a rail tank- er containing 20 t of oil (c_(oil)=2.3kJ/(kgcdot deg)) at 15^circ C The heat loss to the environment is 8%

To determine the power of the electric heater needed to heat the oil, we can use the formula:<br /><br />\[ Q = mc\Delta T \]<br /><br />where:<br />- \( Q \) is the heat energy required,<br />- \( m \) is the mass of the oil,<br />- \( c \) is the specific heat capacity of the oil,<br />- \( \Delta T \) is the change in temperature.<br /><br />Given:<br />- \( m = 20 \) t = 20,000 kg (since 1 t = 1,000 kg),<br />- \( c_{oil} = 2.3 \) kJ/(kg·°C),<br />- Initial temperature \( T_i = 15^{\circ}C \),<br />- Final temperature \( T_f = 65^{\circ}C \),<br />- Heat loss to the environment = 8%.<br /><br />First, calculate the change in temperature:<br /><br />\[ \Delta T = T_f - T_i = 65^{\circ}C - 15^{\circ}C = 50^{\circ}C \]<br /><br />Next, calculate the heat energy required:<br /><br />\[ Q = mc\Delta T = 20,000 \, \text{kg} \times 2.3 \, \text{kJ/(kg·°C)} \times 50^{\circ}C = 2,300,000 \, \text{kJ} \]<br /><br />Since there is an 8% heat loss, the actual heat energy supplied by the heater should be:<br /><br />\[ Q_{\text{actual}} = \frac{Q}{1 - \text{heat loss percentage}} = \frac{2,300,000 \, \text{kJ}}{1 - 0.08} = \frac{2,300,000 \, \text{kJ}}{0.92} = 2,500,000 \, \text{kJ} \]<br /><br />Now, convert the heat energy from kJ to Joules:<br /><br />\[ Q_{\text{actual}} = 2,500,000 \, \text{kJ} \times 1,000 \, \text{J/kJ} = 2,500,000,000 \, \text{J} \]<br /><br />Finally, calculate the power of the heater in watts (1 watt = 1 J/s):<br /><br />\[ \text{Power} = \frac{Q_{\text{actual}}}{\text{time}} = \frac{2,500,000,000 \, \text{J}}{3600 \, \text{s}} = 694,444.44 \, \text{W} \]<br /><br />Therefore, the power of the electric heater needed to heat the oil is approximately 694.44 kW.