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Problem Set 11 Solutions

# Problem Set 11 Solutions - Problem Set XI Solutions Fall...

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Problem Set XI Solutions Fall 2006 Physics 200a 1. How much heat is needed to convert 1 kg of ice at -10 o C to steam at 100 o C ? Remember ice and water do not have the same specific heat. In general the heat necessary to warm a material that doesn’t change phase is: Q = m c T ! ! " Where here the temperature change is in Kelvin, the specific heat c is in J / kg ! K , and the mass is in kg . Also in general, the energy required to change the phase of a mass m with heat of transformation L is: Q = m L ! The heat necessary to warm the ice to 0 o C is: Q ice = # \$ # \$ kg K kg J K 1 2050 10 % % & ( ( ) * ! = 20.5 kJ We can look up the heat of fusion of water to get: Q ice + water = # \$ kg kg kJ 1 334 % % & ( ( ) * = 334 kJ The heat then necessary to warm the water from 0 o C to 100 o C is: Q water = # \$ # \$ kg K kg J K 1 4184 100 % % & ( ( ) * ! = 418.4 kJ And finally, we look up the heat of vaporization of water getting: Q water + steam = # \$ kg kg kJ 1 2257 % % & ( ( ) * = 2257 kJ So, the total heat needed is: Q total = Q ice + Q ice + water + Q water + Q water + steam Q total = 20.5 kJ + 334 kJ + 418.4 kJ + 2257 kJ , 3030 kJ 2. If 400 g of ice at -2 o C is placed in 1 kg of water at 21 o C what is the end product when equilibrium is reached? First we need to determine if the ice will completely melt. To do this we find the heat necessary to warm the ice to 0 o C and then to melt it with the same method as in Prob. 1. Q ice = # \$ # \$ kg K kg J K 4 . 2050 2 % % & ( ( ) * ! = 1.64 kJ Q ice + water = # \$ kg kg kJ 4 . 334 % % & ( ( ) * = 133.6 kJ Q melt = Q ice + Q ice + water = 135.24 kJ The maximum amount of heat that the water can give before beginning to freeze is:

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Q water = # \$ # \$ kg K kg J K 1 4184 21 % % & ( ( ) * ! = 87.864 kJ Since this is smaller than the amount of heat necessary to completely melt but far more than the amount necessary to warm the ice to 0 o C we now know that the water will warm all of the ice to 0 o C and melt some fraction of the ice. We already know that 1.64 kJ is necessary to warm the ice to 0 o C , so the remaining 86.224 kJ available from the water will go toward melting the ice. The amount of ice melted is then: f L Q m - kg kJ kJ 334 224 . 86 - , 258 g Thus we’re left with about 1.258 kg of water and 0.142 kg of ice, all at 0 o C. 3. To find c x , the specific heat of material X , I place 75 g of it in a 30 g copper calorimeter that contains 65 g of water, all initially at 20 o C . When I add 100 g of water at 80 o C , the final temperature is 49 o C . What is c x ? Here we want to make use of conservation of energy, namely the heat absorbed by the original amount of material X , the calorimeter and the initial water is equal to the heat lost by the added hot water. This gives us the equation: ( m x c x + m copper c copper + m water c water ) ! t initial.components = m hot.water c hot.water ! t hot.water where both temperature changes are positive numbers.
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