PHYS 102 HOMEWORK VIII

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1 PHYS 102 HOMEWORK VIII DUE DATE: 24/12/2010 FRIDAY 11:50 (end of class) Please Sign Below I pledge my honor that I have not copied the solutions from a solution manual or from a friend. I have neither given nor received assistance. NAME: SIGNATURE: UNSIGNED HOMEWORKS WILL NOT BE ACCEPTED!

2 Problem 1 [20 pts] Q = 50 cal and W = 20 cal, when a system is taken from state i to state f along path iaf. Along path ibf, Q = 36 cal. (a) What is W along path ibf? (b) If W = -13 cal for the return path fi, what is Q for this path? (c) If E int,i = 10 cal, what is E int,f? (d) If E int,b = 22 cal, what is Q for path ib and (e) path bf? Notice that W = 20 cal is the work done by the system. So the work done on the system is W on the system = 20 cal. (a) The change in internal energy ΔE int is the same for path iaf and path ibf. According to the first law of thermodynamics, ΔE int = Q + W on the system, where Q is the heat absorbed and W on the system is the work done on the system. Along iaf Along ibf, ΔE int = Q + W on the system = 50 cal 20 cal = 30 cal. W on the system = ΔE int Q= 30 cal 36 cal = 6.0 cal. (b) Since the curved path is traversed from f to i the change in internal energy is 30 cal and Q = ΔE int + W = 30 cal 13 cal = 43 cal. (c) Let ΔE int = E int, f E int, i. Then, E int, f = ΔE int + E int, i = 30 cal + 10 cal = 40 cal. (d) The work W bf for the path bf is zero, so Q bf = E int, f E int, b = 40 cal 22 cal = 18 cal. (e) For the path ibf, Q = 36 cal so Q ib = Q Q bf = 36 cal 18 cal = 18 cal.

3 Problem 2 [20 pts] An amount of gas within a chamber passes through the cycle shown in the figure. Determine the energy transferred by the system as heat during process CA if the energy added as heat Q AB during process AB is 20.0 J, no energy is transferred as heat during process BC, and the net work done during the cycle is 15.0 J. Since the process is a complete cycle (beginning and ending in the same thermodynamic state) the change in the internal energy is zero and the heat absorbed by the gas is equal to the work done by the gas: Q = W by the system. In terms of the contributions of the individual parts of the cycle Q AB + Q BC + Q CA = W by the system and Q CA = W by the system Q AB Q BC = 15.0 J 20.0 J 0 = 5.0 J. This means 5.0 J of energy leaves the gas in the form of heat.

4 Problem 3 [20 pts] The cross section of a wall are made of three layers. The thicknesses of the layers are L 1, L 2 = 0.700L 1, and L 3 = 0.350L 1. The thermal conductivities are k 1, k 2 = 0.900k 1, and k 3 = 0.800k 1. The temperatures at the left and right sides of the wall are T H = 30.0 o C and T C = o C, respectively. Thermal conduction is steady. (a) What is the temperature difference ΔT 2 across layer 2 (between the left and right sides of the layer)? If k 2 were, instead, equal to l.1k 1, (b) would the rate at which energy is conducted through the wall be greater than, less than, or the same as previously, and (c) what would be the value of ΔT 2? (a) We take the rate of conductive heat transfer through each layer to be the same. Thus, the rate of heat transfer across the entire wall P w is equal to the rate across layer 2 (P 2 ). Using the equation for P and canceling out the common factor of area A, we obtain T H T c (L 1 /k 1 + L 2 /k 2 + L 3 /k 3 ) = ΔT 2 (L 2 /k 2 ) 45 C (1 + 7/9 + 35/80) = ΔT 2 (7/9) which leads to ΔT 2 = 15.8 C. (b) We expect (and this is supported by the result in the next part) that greater conductivity should mean a larger rate of conductive heat transfer. (c) Repeating the calculation above with the new value for k 2, we have 45 C (1 + 7/ /80) = ΔT 2 (7/11) which leads to ΔT 2 = 13.8 C. This is less than our part (a) result which implies that the temperature gradients across layers 1 and 3 (the ones where the parameters did not change) are greater than in part (a); those larger temperature gradients lead to larger conductive heat currents (which is basically a statement of Ohm s law as applied to heat conduction ).

5 Problem 4 [20 pts] Ethyl alcohol has a boiling point of 78.0 o C, a freezing point of o C, a heat of vaporization of 879 kj/kg, a heat of fusion of 109 kj/kg, and a specific heat of 2.43 kj/kg.k. How much energy must be removed from kg of ethyl alcohol that is initially a gas at 78.0 o C so that it becomes a solid at -114 o C? To accomplish the phase change at 78 C, Q = L V m = (879 kj/kg) (0.510 kg) = kj must be removed. To cool the liquid to 114 C, Q = cm ΔT = (2.43 kj/ kg K ) (0.510 kg) (192 K) = kj, must be removed. Finally, to accomplish the phase change at 114 C, Q = L F m = (109 kj/kg) (0.510 kg) = kj must be removed. The grand total of heat removed is therefore ( ) kj = 742 kj.

6 Problem 5 [20 pts] (a) Two 50 g ice cubes are dropped into 200 g of water in a thermally insulated container. If the water is initially at 25 o C, and the ice comes directly from a freezer at -15 o C, what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used? (a) We work in Celsius temperature, which poses no difficulty for the J/kg K values of specific heat capacity since a change of Kelvin temperature is numerically equal to the corresponding change on the Celsius scale. There are three possibilities: None of the ice melts and the water-ice system reaches thermal equilibrium at a temperature that is at or below the melting point of ice. The system reaches thermal equilibrium at the melting point of ice, with some of the ice melted. All of the ice melts and the system reaches thermal equilibrium at a temperature at or above the melting point of ice. First, suppose that no ice melts. The temperature of the water decreases from T Wi = 25 C to some final temperature T f and the temperature of the ice increases from T Ii = 15 C to T f. If m W is the mass of the water and c W is its specific heat then the water rejects heat Q = c m ( T T ). W W Wi f If m I is the mass of the ice and c I is its specific heat then the ice absorbs heat Q= c m ( T T ). I I f Ii Since no energy is lost to the environment, these two heats (in absolute value) must be the same. Consequently, c m ( T T ) = c m ( T T ). W W Wi f I I f Ii The solution for the equilibrium temperature is

7 T f Chapter 20 [HOMEWORK 8] Return by Friday, December 24 c m T = c m + c m T + c m W W Wi I I Ii W W I I (4190J / kg K)(0.200kg)(25 C) + (2220J/kg K)(0.100kg)( 15 C) = (4190 J/kg K)(0.200 kg) + (2220 J/kg K)(0.100 kg) = 16.6 C. This is above the melting point of ice, which invalidates our assumption that no ice has melted. That is, the calculation just completed does not take into account the melting of the ice and is in error. Consequently, we start with a new assumption: that the water and ice reach thermal equilibrium at T f = 0 C, with mass m (< m I ) of the ice melted. The magnitude of the heat rejected by the water is Q = cwmwt Wi, and the heat absorbed by the ice is Q= c m (0 T ) + ml, I I Ii F where L F is the heat of fusion for water. The first term is the energy required to warm all the ice from its initial temperature to 0 C and the second term is the energy required to melt mass m of the ice. The two heats are equal, so cwmwtwi = c. ImT I Ii + mlf This equation can be solved for the mass m of ice melted: c m T + c m T m = L W W Wi I I Ii F (4190J / kg K)(0.200kg)(25 C) + (2220J / kg K)(0.100kg)( 15 C ) = = = kg 53g J / kg Since the total mass of ice present initially was 100 g, there is enough ice to bring the water temperature down to 0 C. This is then the solution: the ice and water reach thermal equilibrium at a temperature of 0 C with 53 g of ice melted.

8 (b) Now there is less than 53 g of ice present initially. All the ice melts and the final temperature is above the melting point of ice. The heat rejected by the water is Q = c m ( T T ) W W Wi f and the heat absorbed by the ice and the water it becomes when it melts is Q= c m (0 T ) + c m ( T 0) + m L. I I Ii W I f I F The first term is the energy required to raise the temperature of the ice to 0 C, the second term is the energy required to raise the temperature of the melted ice from 0 C to T f, and the third term is the energy required to melt all the ice. Since the two heats are equal, c m ( T T ) = c m ( T ) + c mt + m L. W W Wi f I I Ii W I f I F The solution for T f is T f = c m T + c mt m L W W Wi I I Ii I F c ( m + m ) W W I. Inserting the given values, we obtain T f = 2.5 C.

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