Questions 1 - 4 are considered fundamental problems. They are at the level of a
ID: 1528511 • Letter: Q
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Questions 1 - 4 are considered fundamental problems. They are at the level of a mid to advanced exam question. Most of these questions are from previous exams. An oxygen molecule falls in a vacuum. From what height must it fall so that its kinetic energy at the bottom equals the average energy of an oxygen molecule at 800 K? The mass of one O_2 molecule is 5.312 times 10^-26 kg. In the summer of 2010 a huge piece of ice roughly four times the size of Manhattan and 500 m thick caved off the Greenland mainland. (a) How much heat would be required to melt this iceberg (assumed to be at 0 degree C) into liquid water at 0 degree C? The annual U.S. energy consumption is 1.2 times 10^20 J. (b)) If all the U.S.'s energy was used to melt the ice, how many days would it take to do so? A 0.15-kg of aluminum contains 0.2 kg of water. Initially, the water and the cup have a common temperature of 18 degree C. An unknown material (m = 0.04 kg) is heated to a temperature of 97 degree C and then added to the water. The temperature of the water, the cup and the unknown materials is 22 degree C after thermal equilibrium is re-established. Ignoring the small amount of heat gained by the thermometer, find the specific heat capacity of the unknown material. Beer brewing begins with steeping grains in hot water, releasing the sugars inside. The sugar water is then heated to a boil and hops added. The hot temperature of the boil extracts oils from the hops and provides sanitation from unwanted bacteria. The whole mess is cooled down and once it is safe enough for yeast to survive, they are added. They yeast convert the sugars to alcohol, and the oils from the hops provide many things including flavor and antibacterial.Explanation / Answer
SOL 1. the average energy of a diatomic gas like oxygen is 5/2 * k * T , where k = boltzmann's constant
and T = temperature in kelvin.
so here Eavg. = 5/2 * (1.380 * 10-23 m2 -kg s-2 K-1) * (800 K) = 2.7612 * 10-20 = (mass of O2) * (gravity ,g) * (height ,h)
i.e.( 2.7612 * 10-20) / (5.312 * 10-26 * 9.81 ) = 52987.17
so height, h = 52987.17 m.
SOL. 2. heat required for melting of iceberg = (mass of iceberg in kg) * (latent heat of iceberg at 0 degree celcius)
= ( density of iceberg * volume of iceberg ) * (latent heat )
where density of ice = 917 kg/m3 , volume of iceberg = 4* 59.1*106 * 500 = 1.182 * 1011 m3 ( considering area of manhattan = 59.1 km2 = 59.1 * 106 m2 )
therefore total mass of iceberg = 1.083894 * 1014 kg.
(a). heat required = 1.083894 * 1014 * 334 = 3.62 * 1016 joules.
(b). if annual heat consumption of US is 1.2 * 1020 joules ( considering 1 year = 365 days)
or per day consumption of US is = (1.2 * 1020) / 365 = 3.288 * 1017 joules / day.
so number of days required = (total heat required to melt the iceberg) / ( per day consumption of US)
= (3.62 * 1016) / (3.288 * 1017) = 0.110 days.
or less than 1 day.
SOL 3). taking mass of aluminium = mAL kg , specific heat = cAL J/(kg-C) =900
mass of water = mw kg , specific heat = cw ( J/kg- C) =4186
mass of unknown material = mun kg , specific heat = cun J/(kg-C)
initial temperature of aluminium and water Ti = 18o C and that of material Tun = 97o C
the final temperature of all three are Tf = 220 C
so here HEAT GAINED BY (ALUMINIUM + WATER) = HEAT LOSS BY UNKNOWN MATERIAL
ie. mAL * cAL * ( 22-18) + mw * cw * (22-18) = mun * cun * ( 97-22)
therefore Cun = (0.15 * 900 * 4 + 0.2*4186*4 ) / (0.04 * 75)
= 1296.267 J/ (kg-C)
SOL. 4 . question is incomplete.
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