HCI Assignment 10 Answer Thermal properties of matter

Thermal Properties of Matter 1 Name: Index No: Date: Class: Assignment 10: Thermal properties of matter 1. A student wants to investigate the relationship between the amount of energy absorbed by water and its subsequent increase in temperature. He observed that a certain amount of water absorbed 10 .0 kJ and its temperature increased by 5.0 oC. He then concluded that "the specific heat capacity of water is 2000 J oC-1 ". However, he found in a textbook that the correct value for the specific heat capacity of water to be 4200 J kg-1 oC-1. (a) Explain why the student's conclusion is incorrect. [1] He did not account for the mass of water in his calculation of the specific heat capacity. [1] ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. (b) State the name of the quantity the student had calculated instead. Give the correct unit for this quantity. [1] Heat capacity. J oC-1 or J /oC [1] ……………………………………………………………………………………………………………….. (c) Determine the mass of water the student used, in kg. [1] Q = mc() 10 000 J = m (4200 J kg-1 oC-1)(5.0 oC) m = 0.48 kg (to 2 s.f.) [1] 2. A chef pours 2 .00  102 g of water at a temperature of 15.0 C into a hot aluminium saucepan with a mass of 2.50  102 g and a temperature of 120.0 C. Determine the temperature, , of the water and the saucepan when thermal equilibrium is reached . Assume no energy is transferred to or from the surroundings. Take the specific heat capacity of water = 4 200 J/(kg K) and the specific heat capacity of aluminium = 0.900 kJ/(kg K). [2] Q gained by water = Q lost by saucepan mw cw (w) = ms cs (s) (0.200 kg) (4200 J/(kg K)) ( – 15.0 C) = (0.250 kg) (900 J/(kg K)) (120.0 C – ) [1]  = 37.2 oC (to 3 s.f.) [1] [accept 37 oC] Note: Value of  calculated is in oC, which is consistent with the temperatures given in the question. There is no need to convert temperatures in oC to K even though s.h.c. is given in J/(kg K). Second mark will not be awarded if 37.2 K is written instead of 37.2 oC. HWA CHONG INSTITUTION Sec 3 Physics

Thermal Properties of Matter 2 3. When a lead ball is dropped vertically on a hard surface, it is deformed upon impact. Assuming all the energy in the kinetic store is transferred to the internal store in the ball upon impact, from what height, h, above the surface, must the ball be dropped to raise its temperature by 1.0 oC? Take the specific heat capacity of lead = 130 J/(kg K) and the gravitational acceleration, g = 10 m/s 2. Neglect air resistance. [2] K.E. of ball just before impact = Q gained to raise its temperature (i.e. increase internal energy) m g (h) = m c () (10 m/s2)(h) = (130 J/(kg K))(1.0 oC) [1] h = 13 m (to 2 s.f.) [1] 4. A lead bullet of mass 3.00  10-3 kg is fired horizontally from a gun such that it hits a steel plate. The speed of the bullet just before impact is 4.00  102 m s-1. Upon hitting the plate, 18 % of the energy in the kinetic store is transferred to the internal store of the bullet. If the specific heat capacity of the bullet is 130 J kg-1 K-1, calculate the rise in temperature of the bullet immediately after impact. [3] K.E. = ½ (3.00  10-3 kg)(4.00  102 m s-1)2 [1] = 240 J Energy gained upon impact = 18/100 X 240 J [1] = 43.2 J Q = mc() 43.2 J = (3.00  10-3 kg)(130 J kg-1 K-1)()  = 110 oC (to 2 s.f.) [1] [accept 110 K. Note: use symbol Δ, not . ] 5. A 1.00 kg copper block is placed at the top of a ramp as shown in the diagram below. When released from rest, the block slides down the rough ramp and reaches the bottom with a speed of 3.50 m s-1. Take gravitational acceleration, g = 10 m s-2. (a) Calculate the amount of energy transferred to the internal store of the block -ramp-earth system, due to work done against friction, when the block reaches the bottom of the ramp. [2] K.E.initial + G.P.E.initial = K.E.final + G.P.E.final + W against friction 0 + (1.00)(10)(2.00) = ½ (1.00)(3.50)2 + 0 + W against friction [1] W against friction = 13.875 = 13.9 J (to 3 s.f.) [1] (b) Assume 75 % of the energy calculated in (a) is transferred to the internal store of the block . Determine the specific heat capacity of copper , if the temperature of the block increases by 0.027 oC, [2] Q = mc() 0.75 x 13.875 J = (1.00 kg)(c)(0.027 oC) [1] c = 3.9  102 J kg-1 K-1 (to 2 s.f.) [1] 2.00 m copper block ramp

Thermal Properties of Matter 3 (c) Explain the change in the motion of the atoms in the copper block as its temperature increased. [2] As the block's temperature increased, the average kinetic energy of its atoms increased. [1] The atoms vibrated more vigorously about their mean (fixed) positions. [1] [reject: ‘move faster’; use ‘energy’, not ‘thermal energy’] ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. 6. Wet clothes dry faster when they are placed outdoors in the sun on a hot, windy day as compared to when placed indoors without any wind. (a) Explain, in terms of molecules, how the h ot outdoor environment allows the wet clothes to dry faster. [2] On a hot day, the temperature of water in the wet clothes placed outdoors is higher, thus the average kinetic energy of the water molecules increases. [1] More molecules near surface of wet clothes have sufficient energy to overcome intermolecular forces (attraction) and escape into the air, thus increasing the rate of evaporation of water. [1] [Intention of question is to explain evaporation, in terms of molecules, thus explanation of ‘water evaporates faster’ or ‘rate of evaporation increases’ is rejected. reject: water molecules evaporate faster; temperature of water molecules increases] ……………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………….. (b) Explain, in terms of molecules, how the windy outdoor environment allows the wet clothes to dry faster. [2] On a windy day, (water) vapour molecules formed by evaporation (of water) near surface of wet clothes are carried away at a faster rate. [1] More molecule s e scape int o air, without colliding into other gaseous molecules (or without bouncing back into water), thus increasing the rate of evaporation of water. [1] ……………………………………………………………………………………………………………….. ………………………………………………………………………………………………………………..

Thermal Properties of Matter 4 7. The diagram shows a graph of temperature, T, against time for 15.0 g of water when it was placed in a freezer. The water is initially at 44.0 oC. (a) Calculate the amount of energy removed from the water in t2 s. Take the specific heat capacity of water, cwater = 4200 J kg-1 K-1 and the specific latent heat of fusion of ice, lf = 336 kJ kg-1. Assume the water became entirely frozen in t2 s. [3] Q removed to change water temperature from 44.0 oC to 0 oC = m c () = (0.0150 kg)(4200 J kg-1 K-1)(44.0 oC – 0 oC) [1] = 2772 J Q required to change water at 0 oC to ice at 0 oC = m lf = (0.0150 kg)(336  103 kJ kg-1) [1] = 5040 J Total Q removed = 2772 + 5040 = 7812 J = 7800 J (to 2 s.f.) [1][accept: 7810 J] (b) Given the freezer removed energy at a constant rate of 220 J s -1, determine the value of t1 and t2. [3] t1 = 2772 / 220 = 12.6 s = 13 s (to 2 s.f.) [1] t2 - t1 = 5040 / 220 = 22.909 s [1] t2 = 35.5 s = 36 s (to 2