Unformatted text preview: or notes that the 8.2g lead bullet that was stopped in a doorframe melted completely on impact. Assume the bullet was red at room temperature (20 C). The heat of fusion for lead is 5.9 kcal/kg and its melting point is 327 C. The speci c heat of lead is 0.031 cal/g* C or 130 J/kg C (a) 3 points] What is the heat required to get to the melting point? Heat = 8:2g (327 ; 20) C 0:031cal=g C = 78 cal (b) 3 points] What is the heat required for melting the bullet? Heat melting = 8:2g 5:9 103 10;3 = 48.4 cal (c) 2 points] What is the energy in joules required to heat and melt the bullet? Energy = 4:18 (78 + 48:4) j = 528 joules (d) 2 points] What does the investigator calculate was the minimum muzzle velocity of the gun?
1 KE = 2 mv2 = 528 joules. v = 2 528=8:2 10;3 = 359m=s p 1 5. 15 points] You plan to have iced tea for your afternoon party. You have an insulated container with 3 kg of tea (essentially water) at 20 C, to which you add ice at 10 C. (a) 5 points] How much ice at 10 C do you need to add to the 20 C tea in order to have a resulting mixture of tea (water) and 0.2 kg of ice in equilibrium? What is the equilibrium temperature? The equilibrium temperature will be 0 since we have ice and water together. The ice will warm and all but 0.2 kg will melt. (We will assume the speci c heat capacity of ice to be the same as water in reality it is about half.) The heat lost by the water is Qw = mw cw Tw = 3000g 1cal=g C ;20 C = ;6 104cal The heat gained by the ice is Qice = micecice Tice + (mice ; 200g)Lf = mice 1cal=g C 10 C + (mice ; 200g) 80cal=g = (10 + 80)micecal=g ; 1:6 104cal The two heats have to sum to zero so we have (10 + 80)micecal=g ; 1:6 104cal = 6 104cal or 90mice = 7:6 104g yielding mice = 844g = 0:844kg (b) 5 points] Suppose the container of the iced tea is left in the state equilibrium describe in part (a) and the temperature outside the container is xed at 20 C. Suppose further that the container has a total surface area of 2500 cm2, a thickness of 1 cm and is made of material with thermal conductivity 0.025 W/m C. While there is still ice left, nd the rate at which heat energy is entering the container and how long it takes the ice to melt. We use Newton's heat conductivity equation dQ = ; A T = 0:025W=m C 2500cm2 20 C = 0:025 25 20W = 12:5W = 3cal=s dt l 1cm At the end...
View
Full Document
 Fall '08
 Packard
 Thermodynamics, Energy, Heat, Heat engine, Carnot heat engine

Click to edit the document details