#$&* course Phy 122 Your solution, attempt at solution:If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: ** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is directly proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as • `dQ = mass * specific heat * `dT. (General College and University Physics students note that most substances do not quite behave in this ideal fashion; for most substances the specific heat is not in fact strictly constant and for most substances changes with temperature.) For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation • m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently • m1 c1 `dT1 = - m2 c2 `dT2. That is, whatever energy one substance loses, the other gains. In this situation we know the specific heat of water, the two temperature changes and the two masses. We can therefore solve this equation for specific heat c2 of the unknown substance. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 86C+273= 359K; -100C+273=173K; 5500C+273=5773K 5/9(78-32)=26C+273=299K and 5/9(-459-32)= -273+273=0K confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The Kelvin temperature is 273 K higher than the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree). • 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -100 + 273 K = 173 K, (5500 + 273) K = 5773 K. The freezing point of water is 0 C or 32 F, and a Fahrenheit degree is 5/9 the size of a Celsius degree. Therefore • 78 F is (78 F - 32 F) = 46 F above the freezing point of water. • 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing. • Since freezing is at 0 C, this means that the temperature is 26 C. • The Kelvin temperature is therefore (26 + 273) K = 299 K. Similar reasoning can be used to convert -459 F to Celsius • -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing. • This is -273 C or (-273 + 273) K = 0 K. • This is absolute zero, to the nearest degree. Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Temperature increased by factor of 40/9 P1V1/T1=P2V2/T2 T2= (P2/P1)*(V2/V1)*T1 T2=40*1/9*273=1302K confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: First we reason this out intuitively: If the air was compressed to 1/9 its original volume and the temperature didn’t change, it would end up with 9 times its original pressure. However the pressure changes from 1 atm to 40 atm, which is a 40-fold increase. The only way the pressure could end up at 40 times the original pressure, as opposed to 9 times the original, would be to heat up. Its absolute temperature would therefore have to rise by a factor of 40 / 9. Its original temperature was 20 C = 293 K, so the final temperature would be 293 K * 40/9, or over 1300 K. Now we reason in terms of the ideal gas law. P V = n R T. In this situation the number of moles n of the gas remains constant. Thus P V / T = n R, which is constant, and thus P1 V1 / T1 = P2 V2 /T2. The final temperature T2 is therefore • T2 = (P2 / P1) * (V2 / V1) * T1. From the given information P2 / P1 = 40 and V2 / V1 = 1/9 so • T2 = 40 * 1/9 * T1. The original temperature is 20 C = 293 K so that T1 = 293 K, and we get • T2 = 40 * 1/9 * 293 K, the same result as before. Your Self-Critique: OK Your Self-Critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!