QUESTION 1: 11 An electric fumace is used to melt 10kg of aluminium with an init
ID: 2293002 • Letter: Q
Question
QUESTION 1: 11 An electric fumace is used to melt 10kg of aluminium with an initial temperature of 20°C Take the melting point of aluminium to be 660°C, c 950Jkg K and the slhf 387kJ/kg Determine the power (heat energy) required to accomplish the conversion in 20 minutes. Assume the efficiency of the fumace to be 75 percent. A current of 5 amperes is sustained in a 1 3? resistor for 6 minutes The resistor is placed inside 0 44 ltres of water in a container The water equrvalent of the vessel and heater is 0.017kg Calculate the heat generated in kilojoule, (5) 1 2 And detemine the nse in temperature of the water. ignore any loss of heat and assume the specific heat capacity of water to be 4190 Joule per kilogram Kelvn 13 ,L/kg. K)]. [151Explanation / Answer
Q. Amount of heat required to melt the Al = amount of heat required to bring the Al to 660C + amount of heat required to melt Al at 660C = mc delta T + m L = 10*950* ( 933-293) + 10*378*1000 ( since 660C in Kelvin is 660+273 = 933K and 20C in Kelvin scale is 20+273 = 293K )
= 6080X103? + 3780X10?3? = 9860 kJ
Q. Power = I2?R= 25*1.3 = 32.5W => Energy = Power * time = 32.5* 6 *60 = 11.7kJ
Heat generated = 11.7kJ
Q. 1 liter of water = 1kg => 0.44liters = 0.44kg
The rise in temperature : (0.44 + 0.017 ) * 4190 * (delta T ) = 11.7 x 103? => delta T = 6.11 K