A) if the water has a mass of 50kg, how massive does the aluminum need to be so that the final temperature is 100degrees C ?B) if the aluminum was actually 500g, how much water would be boiled of?
(A) Let the aluminum is of m kg, than :- =Q(gain) by water = Q(loss) by aluminum =m x s(water) x delta t*C = m x s(aluminum) x delta t*C =50 x 4187 x (100 - 50) = m x 897 x (220 - 100) =m = 97.25 kg (B) Let the amount of water boiled off is m kg =Q(loss) by aluminum = Qwater(50*C to 100*C) + Q100*C(water to vapor) =500 x 10^-3 x 897 x (220 - 100) = 50 x 4187 x (100 - 50) + m x 2257 x 10^3 =m x 2257 x 10^3 = -10413680 =No water will be boiled off as the energy provided by aluminum is less than required to raise the temperature of water to 100*C.
A chunk of aluminum is at 220degees C is thrown into water that is initially 50degrees C? The specific heat of Al = 903 J/(kg * ?C) The specific heat of Water = 4180 J/(kg * ?C) Heat of vaporization of water = 2.26 * 10^6 J/kg A) if the water has a mass of 50kg, how massive does the aluminum need to be so that the final temperature is 100degrees C ? Heat = mass * ?T * specific heat Heat gained by water = 50 * (100 – 50) * 4180 Heat lost by aluminum = M * (220 – 100) * 903 Heat lost must equal heat gained M * (220 – 100) * 903 = 50 * (100 – 50) * 4180 M * 120 * 903 = 50 * 50 * 4180 M = [50 * 50 * 4180] ÷ [120 * 903] M = 96.44 Kg B) if the aluminum was actually 500g, how much water would be boiled of? The heat energy required to raise the temperature of 50 kg of water from 50 ?C to 100 ?C = [50 * 50 * 4180] Joules of energy [50 * 50 * 4180] Joules of energy will cool 0.500 kg of Aluminum how many degrees? 0.500 * ?T * 903 = [50 * 50 * 4180] ?T = [50 * 50 * 4180] ÷ [0.500 * 903] ?T = 23,145.07 ?C I assume the mass of aluminum = 500 kg, not 500g 500 * ?T * 903 = [50 * 50 * 4180] ?T = [50 * 50 * 4180] ÷ [500 * 903] ?T = 23.15 ?C So the temperature of the 500 kg of Al decreased 23.15 ?C. Temperature of Al = 196.85 ?C To boil 1 kg of water, requires 2.26 * 10^6 Joules of energy To boil M kg of water, requires (M * 2.26 * 10^6) Joules of energy The 500 kg of Al can cool from 220 ?C to 100 ?C as it boils the water. When the 500 kg of Al cools 100? C, it releases (500 * 120 * 903 ) = 5.418 * 10^7 Joules of heat energy M * 2.26 * 10^6 = 5.418 * 10^7 M = 23.97 kg of water was boiled into steam at 100 ?C The temperature of the remaining 26.03 Kg of water was 100 ?C
