DHS 12 First Law of Thermodynamics (Tutorial Solutions)

Dunman High School 2024 Physics H2 Tutorial Topic 9: First Law of Thermodynamics For Internal Use Only lim.boonsiong@dhs.edu.sg 1 Suggested Answers 1. 10.0 g of ice at –10 oC are introduced into 100 g of water at 50 oC contained in a copper calorimeter with a mass of 50.0 g. Calculate the final temperature reached. (specific latent heat of ice = 0.34 × 106 J kg−1 specific heat capacity of ice = 2.09 × 103 J kg−1 K−1 specific heat capacity of copper = 380 J kg−1 K−1 specific heat capacity of water = 4200 J kg−1 K−1) [38.1oC]    o oheat abs b A orbed ( by i e ce h fr e o u m ssume the heat exchang wit th s rr 10 C to 0 C by i u i ce to m o nding is negligible. Let the f elt by wate r r from melted ice t e e o reac nal temp ratu ( t be u . hea s p i h pl ) ed                                                     3 3 3 6 3 2 3 2 o 3 3 10.0 10 2.09 10 10 10.0 10 0.34 10 10.0 10 4200 0 2.09 10 3.40 10 42 4.39 10 50 38.13 C y water to reach by calorimeter to reach ) 100 10 4200 50 50.0 10 380 50         2. 1 kg of vegetables, having a specific heat capacity 2200 J kg−1 K−1, at a temperature of 373 K are plunged into a mixture containing some ice and 1 kg of water at 273 K. After all the ice has melted, the final temperature of the entire mixture is 300 K. Calculate the mass of ice originally present. Assume no heat loss to the container and the surroundings. Specific latent heat of fusion of ice = 3.34×105 J kg−1 [0.105 kg] (Assuming the ice point is at 273 K.)            5 Let the mass of ice be kg. heat absorbed by (ice to melt water from melted ice 1 kg of water) 3.34 10 1 4200 300 273 0.105 kg heat supplied by vegetables 1 2200 373 300 m m m m           

Dunman High School 2024 Physics H2 Tutorial Topic 9: First Law of Thermodynamics For Internal Use Only lim.boonsiong@dhs.edu.sg 2 3. A thermally insulated vessel containing liquid water and water vapour is connected to a vacuum pump which removes water vapour continuously. When the temperature reaches 0 oC, the vessel contains 110 g of liquid water. What mass of ice has been formed when no liquid remains? Specific latent heat of fusion of water = 3.40×105 J kg−1; Specific latent heat of vaporisation of water 1= 2.52×106 J kg−1 (at 0.01 oC) [96.9 g] Let mice (in kg) be the mass of the ice formed, then the mass of vapor mvapor produced would be (0.110  mice) kg By conservation of energy, heat given off in freezing water to ice = heat required to boil water close to 0oC       6 5 6 0.110 0.110 0.110 0.110 2.52 10 3.40 10 2.52 10 0.0969 kg 96.9 g ice f vapor v ice v ice f v v v ice f v m L m L m L m L L L Lm L L               4. A lead bullet is fired with a muzzle velocity of 320 m s−1 and becomes immediately embedded in clay. Estimate the fraction of the mass of the bullet which melts, assuming that all the heat generated remains as internal energy in the lead. Given that the muzzle temperature of the bullet is 30 oC, a. melting point of lead is 330 oC, b. the specific heat capacity of lead is 130 J kg−1 K−1 and c. the specific latent heat of fusion of lead is 2.1 × 104 J kg−1. [0.58]        4 2 heat supplied to (increase bullet's temperature melt ' of lead) loss in kinetic en ergy 1130 330 30 ' 2.1 10 320 2 21000 ' 12200 ' 0.581 m m m m m m m m         1 The heat of vaporization diminishes with increasing temperature.

Dunman High School 2024 Physics H2 Tutorial Topic 9: First Law of Thermodynamics For Internal Use Only lim.boonsiong@dhs.edu.sg 3 5. Cooling water enters the heat exchanger in the turbine hall of a nuclear power station at 6.0 oC and leaves at 14.0 oC. The rate of heat removal by the water is 6.7  109 J per minute. The specific heat capacity of water is 4200 J kg−1 K−1. Determine the mass of water flow per unit time. [3320 kg s−1] By conservation of energy, rate of heat gain by water = rate of heat loss by heat exchanger   mc T t m t                  9 9 1 6.7 10 60 6.7 10 1 3320 kg s60 4200 8.0 6. (a) Using the concept of internal energy, compare the internal energy per unit mass of water and water vapour at the same temperature. (a) The internal energy U of a system is determined by the state of the system, i.e. U of system is a function of state coordinates p, V and T. It can be expressed as the sum of a random distribution of kinetic and potential energies associated with the molecules of the system. At the same temperature, both water and water vapour molecules have the same kinetic energy of random motion. When unit mass of water turns into vapour , it does work in separating the molecules, against their mutual attractions. This amount of energy is a measure of the increase in potential energy of the molecules of water in gaseous state over that in the liquid state, at the same temperature. Thus the internal energy per unit mass of water vapour is greater than that of water. (b) Explain, using a simple kinetic model for matter, (i) why the temperature of a pure substance does not change when it experiences a change of state. (ii) the specific latent heat of vaporisation2 of water (Lv = 2.26 × 106 J kg−1) is greater than the specific latent heat of fusion (Lf = 3.34 × 105 J kg−1). 2 The specific latent heat of fusion (or vaporisation) of a material is the energy required to change the state of 1 kg of the material from solid to liquid (or liquid to vapour) without change of temperature. The force between molecules in a gas varies with the distance r between molecules somewhat as shown. When molecules are far apart, the intermolecular forces are very small and usually attractive. As a gas is compressed and its molecules are brought closer together, the attractive forces increase. The intermolecular force becomes zero at an equilibrium spacing r0, corresponding roughly to the spacing between molecules in the liquid and solid states. In liquids and solids, relatively large pressures are needed to compress the substance appreciably.

Dunman High School 2024 Physics H2 Tutorial Topic 9: First Law of Thermodynamics For Internal Use Only lim.boonsiong@dhs.edu.sg 4 (b) (i) During melting and boiling, all the heat energy absorbed by the pure substance is used to change the molecular structures by breaking the bonds and increase the potential energy of the molecules only. The kinetic energy of the molecules does not change, and hence the temperature of substance does not increase. An impure substance is a mixture of several substances, which melt or boil at slightly different temperatures. The heat absorbed is used