A house has well-insulated walls. It contains a volume of 150 m3 of air at 315 K

Question

A house has well-insulated walls. It contains a volume of 150 m3 of air at 315 K. (a) Consider heating it at constant pressure. Calculate the energy required to increase the temperature of this diatomic ideal gas by 1.8

Explanation / Answer

First approach Follow these steps. What is the unknown? Your problem asks "How much energy...?" The units aren't specified, but since you spot 'calories' in the given information, it will probably be easiest to get your answer in calories. What information is given? You know that: the specific heat is 1.0 cal/g°C the mass of water is 25.1 g the temperature change is 72.0°C How is the information in the problem related to the unknown? The relationship between heat and temperature change is heat = mass × specific heat × temperature change Do the math. Now you're ready to do the calculation itself: heat = (25.1 g)(1.0 cal/g°C)(72.0°C) Is the final answer in the correct units? You want only calories. All other units should 'cancel' in the calculation somewhere. Is the size of the answer reasonable? Your reasoning should go something like this: The specific heat means that it takes 1.0 calories of heat to warm 1 g of water up by 1°C. So it should take 25.1 calories to warm up 25.1 g of water by 1°C. And it should take 72.0×25.1 calories to warm up 25.1 g of water by 72.0°C. You can also check the answer by working the problem backwards. For example, take the calories just calculated and divide them by the temperature change and the grams of water; do you get a specific heat of 1.0 cal/g°C 2nd approach **** (a) **** ?H = m Cp ?T = (140 g ) x (0.24 J/° C g) x (295 - 273° C) = 739 J (with 3 sig figs) **** (b) **** ?H = m Cp ?T but moles = mass / mw ----> mass = moles x mw ?H = (moles x mw) Cp ?T = (1.0 moles x 107.9 g/mole) x (0.24 J/° C g) x (1° C) = 26 J (with 2 sig figs) **** (c) **** ?H = m Cp ?T m = ?H / (Cp ?T) m = (1250 J ) / (0.24 J/° C g x (15.3° C - 12.0° C) ) =1578 grams.... since you only have 3 sig figs..... answer is 1.58 kg ******* Edward ****** you wrote b) Q= (47) x 0.24 x (295 - 273 )= the atomic number of Ag is 47. the atomic mass is 108g/mole and it should have been Q= 1.0 mole x (108 g/mole) x 0.24 x (295 - 273 )= ... third approach Step 1: Heat required to raise the temperature of ice from -10 °C to 0 °C Use the formula q = mc?T where q = heat energy m = mass c = specific heat ?T = change in temperature q = (25 g)x(2.09 J/g·°C)[(0 °C - -10 °C)] q = (25 g)x(2.09 J/g·°C)x(10 °C) q = 522.5 J Heat required to raise the temperature of ice from -10 °C to 0 °C = 522.5 J Step 2: Heat required to convert 0 °C ice to 0 °C water Use the formula q = m·?Hf where q = heat energy m = mass ?Hf = heat of fusion q = (25 g)x(334 J/g) q = 8350 J Heat required to convert 0 °C ice to 0 °C water = 8350 J Step 3: Heat required to raise the temperature of 0 °C water to 100 °C water q = mc?T q = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)] q = (25 g)x(4.18 J/g·°C)x(100 °C) q = 10450 J Heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J Step 4: Heat required to convert 100 °C water to 100 °C steam q = m·?Hv where q = heat energy m = mass ?Hv = heat of vaporization q = (25 g)x(2257 J/g) q = 56425 J Heat required to convert 100 °C water to 100 °C steam = 56425 Step 5: Heat required to convert 100 °C steam to 150 °C steam q = mc?T q = (25 g)x(2.09 J/g·°C)[(150 °C - 100 °C)] q = (25 g)x(2.09 J/g·°C)x(50 °C) q = 2612.5 J Heat required to convert 100 °C steam to 150 °C steam = 2612.5 Step 6: Find total heat energy HeatTotal = HeatStep 1 + HeatStep 2 + HeatStep 3 + HeatStep 4 + HeatStep 5 HeatTotal = 522.5 J + 8350 J + 10450 J + 56425 J + 2612.5 J HeatTotal = 78360 J

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