What is the difference between boiling, evaporation and vaporisation?
Evaporation and boiling are two modes of vaporisation a liquid could undergo.Evaporation is a surface phenomenon. It could take place at all temperatures below the boiling point. For understanding evaporation, we need to understand what is vapour pressure. Vapour pressure is the pressure exerted by a liquid’s vapour on the liquid at phase equilibrium at a particular temperature. Phase equilibrium is attained when an equilibrium is attained between the rates of vaporisation and condensation. When the partial pressure of a liquid is less than the vapour pressure at a given temperature, evaporation takes place. For eg, water kept in a dry room at room temperature will continue to evaporate until the partial pressure equals the vapour pressure at room temperature.Boiling is a bulk phenomenon. It takes place when the vapour pressure of a liquid exceeds the environmental pressure (mostly atmospheric pressure). This causes the liquid to continue vaporising so as to strike a phase equilibrium but doesn't take place since no amount of vapor can fill the whole environment with vapor.
Quick Chemistry / Physics question on heat of vap?
heat lost be water = heat gained by CO2 (m Cp dT) H2O = (m x dHsub) CO2 so.. rearrange and solve dHsub CO2 = (m Cp dT H2O) / (m CO2) dHsub CO2 = (500g x 4.184J/g°C x (66°C - 29°C)) / (125kg) = 619 J/kg ******* you sure about that 125kg? 125 kg? 275 lbs of dry ice dumped into 500g of water? 1/2 L of water? how about we try this with 0.125kg CO2.. dHsub CO2 = (500g x 4.184J/g°C x (66°C - 29°C)) / (0.125kg) = 619x10^3 J/kg choice D is the correct answer
I need help with a Physics question. please help?
the specific heat of mercury is .03 cal/g C, and its boiling point is 357C . the specific heat of water is 1cal/g C. it takes 65 calories of energy to vaporize one gram of mercury and 540 calories to vaporize 1 gram of water. if both substances begin at room temp (about 22C), does it take more energy to boil a gram of mercury or a gram of water? please show calculations. Mercury calculations: Water calculation:
How does evaporation cause cooling?
Each molecule in the liquid phase is attracted to the other molecules in the liquid. When a molecule changes phase ( evaporates) it has to break all those intermolecular forces so it can escape. Only the molecules that have sufficient kinetic energy can do this. These high energy molecules leave to the gas phase and the remaining molecules are the ones that did not have enough energy to evaporate. This, the average kinetic energy of the molecules in the liquid left behind is lower that it was before the hot ones leave. So, evaporation “cools” what is left behind because the heat was used up breaking the hotter molecules out into the gas phase.On a hot day you sweat. That sweat evaporates and carries away heat. Your body gave that heat to the sweat ( water) so your body cooled off. This is important as a means by which your body can avoid overheating.
What is evaporation? What are the factors affecting it?
The turning of any liquid into vapour is called Evaporation. The factors affecting the evaporation are:1. Temperature. As the temperature increases, the rate of evaporation also increases.2. Surface Area. As the Surface area increases, the rate of evaporation increases.3. Density. As the density increases, the rate of evaporation decreases.4. Wind Velocity. The velocity of the wind also affects the rate of evaporation. As the velocity of wind increases, the rate of evaporation increases.Reference and further readingsEvaporation - WikipediaWhat are the 6 factors that affect evaporation?
Chemistry question?
What is the enthalpy of vaporization (▲ Hvap, kJ/mol) of a compound that has a vapor pressure of 111 mmHg at 253 K and 0.304 mmHg at 179 K? (R = 8.314 J/(K ▲ mol))
Latent Heat and Expansion?
A person who eats 3195 Cal each day consumes 1.33743*10^7 J of energy in a day. How much water at 100 degrees C could that much energy vaporize? the latent heat of vaporization of water is 2.26*10^6 J/kg answer in units of g
Which one would require the greatest amount of heat?
the problem is finding the various heats, heat of sublimation came from: http://www.usatoday.com/weather/wlatent.... ================= a) melting 3.0 mol of ice at 0"C heat of fusion is 335 J / gram 335J/g ( 54 g H2O) = 18,090 joules ============ b) evaporating 10.0g H20(l) at 298K heat of vaporization = 2260 J/gram 10 grams = 22,600 joules ============ c) melting 2.0 mol of ice at 0"C and heating the resultant H20(l) to 10"C melting = 12,060 Joules heating 36 grams @ 4.18J/g-C ( 10 C)= 1,505 Joules total 13,565 Joules ================== d) subliming 1.0 mol of ice at 0"C heat so sublimation = 2833 J / gram 18grams (2833 J/g) = 50,994 joules your answer: d) subliming 1.0 mol of ice at 0"C