How much energy is needed to boil water? How much energy would it take to boil 1 liter of water?
To calculate this you have to use the specific heat capacity of water which is 4.184J/gram/Celsius. So if you start from room temperature of 21C, and make it to 100C, the boiling point of water, then the temperature change is 79C. The energy needed E=SH times change in temperature times the total mass in grams(Which is 1000gram for a liter of water). That gives you 4.184•79•1000 which is 330,536J, or 330.54KJ. Kaiser T, MD.
At 1 atm, how much energy is required to heat 83.0 g of water(s) at -22.0 Celsius to water(g) at 141.0 Celsius?
You need five computations here: --the amount of energy needed to heat the ice by 22 °C (22 K) --the amount of energy needed to melt the ice --the amount of energy needed to heat the water by 100 °C (100 K) --the amount of energy needed to boil the water --the amount of energy needed to heat the steam by 41 °C (41 K). The specific heat of ice in this temperature range is about 2.05 J/(g·K). The specific enthalpy of fusion of water is 333.55 J/g at 0 °C. The heat capacity of water averages 4.1927 J/(g·K). The heat of vaporization of water is 2257 J/g at 100 °C. The heat capacity of steam at 100 °C is about 2.080 J/(g·K). (83.0 g)(2.05 J/(g·K))(22.0 K) = 3743 J. (83.0 g)(333.55 J/g) = 27 685 J. (83.0 g)(4.1927 J/(g·K))(100.0 K) = 34 800 J. (83.0 g)(2257 J/g) = 187 330 J. (83.0 g)(2.080 J/(g·K))(41.0 K) = 7078 J. Total: 260 640 J (260.6 kJ) -- round this to 261 kJ because only 3 significant figures are justified here.
At 1 atm, how much energy is required to heat 45.0 g of H2O(s) at –18.0 °C to H2O(g) at 133.0 °C?
There are five things to calculate here: --the energy needed to raise the ice to 0°C --the energy needed to melt the ice --the energy needed to raise the water to 100 °C --the energy needed to boil the water --the energy needed to raise the steam to 133 °C. The specific heat capacity of ice at −10 °C is about 2.05 J/(g·K). The specific enthalpy of fusion of water is 333.55 J/g at 0 °C. The specific heat capacity of water is 4.1927 J/(g·K) (average). The heat of vaporization of water at 100 °C is 2256.2 J/g. The heat capacity of steam at 100 °C is about 2.080 J/(g·K). To heat the ice, we need (45.0 g)(2.05 J/(g·K))(18.0 K) = 1660 J. To melt the ice, we need (45.0 g)(333.55 J/g) = 15 010 J. To heat the water, we need (45.0 g)(4.1927 J/(g·K))(100 K) = 18 870 J. To boil the water, we need (45.0 g)(2256.2 J/g) = 101 530 J. To heat the steam, we need (45.0 g)(2.080 J/(g·K))(33.0 K) = 3089 J. Total = 140 160 J. Round this to 140 000 J (140 kJ) because only 3 significant figures are justified here.
How much energy (heat) is required to convert 52.0 g of ice at –10.0°C to steam at 100°C?
The pertinent equations for this problem are: q = mCpΔT where q = quantity of heat (J) m = mass in grams (g) Cp = specific heat capacity (J/g ºC) ΔT = Tf - Ti (final – initial) q = n ΔHfus or q = ΔHvap where q = quantity of heat (J) n = number of moles (grams / molar mass) ΔHfus = molar heat of fusion ΔHvap = molar heat of vaporization We will now calculate the energy of each step: (a) Heat to raise temperature of ice from -10.0°C to 0°C q = mCpΔT q = (52.0 g)(2.09 J/g•°C)(10°C) q = 1086.8 J = 1.09 kJ (b) Heat to melt ice at 0°C q = n ΔHfus q = (52.0 g / 18.02 g/mol) (6.02 kJ/mol) q = 17.38 kJ (c) Heat to raise temperature of water from 0°C to 100°C q = mCpΔT q = (52.0 g)(4.18 J/g•°C)(100°C) q = 21736 J = 21.74 kJ (d) Heat to vaporize water at 100°C q = n ΔHfus q = (52.0 g / 18.02 g/mol) (40.7 kJ/mol) q = 117.45 kJ Total heat = 1.09 + 17.38 + 21.74 + 117.45 = 157.66 kJ Therefore, the answer is c (157.8 kJ). Hope this helps...good luck!
How much energy is required to raise the temperature of 45.0 g of water from 10.5°C to 148.0°C?
Work the problem in 3 steps: 1) Change temperature of 45 g water from 10.5 C to 100 C: q = m c DT = 45 X 4.184 X 89.5 = 16851 J = 16.85 kJ 2) Heat to vaporize 45 g water: 45 g / 18 g/mol = 2.5 mol X 40.7 kJ/mol = 101.75 kJ 3) Heat to raise temperature of steam: q = m c DT = 45 X 2.02 X 48.0 C = 4.36 kJ Add them together to get 123 kJ (3 sig figs because your mass and several other numbers have 3 sig figs).
How much heat energy is required to convert 3 kg of ice at -20 degree Celsius to steam at 140 degree Celsius?
3kg ice at -20C to steam at 140CCp ice = 2.108kJ/kg•KHeat of fusion of ice = 334kJ/kgCp water = 4.187kJ/kg•KHeat of vaporization of water = 2230kJ/kgCp steam = 1.996kJ/kg•KHeating ice to 0C = 3 x 2.108 x 20 = 126.48kJMelting ice to water at 0C = 3 x 334 = 1002kJHeating water from 0C to 100C = 3 x 4.187 x 100 = 1256.1kJVaporizing water at 100C to steam at 100C = 3 x 2230 = 6690.0kJHeating steam from 100C to 140C = 3 x 1.996 x 40 = 239.52kJTOTAL HEAT REQUIRED = 126.48 + 1002 + 1256.1 + 6690 + 239.52 = 9314.1kJ
How much heat energy is required to change 2 kg of ice at 0°C to water at 20°C?
To make your concept clear about such type of questions, you need to understand the concept of latent heat and sensible heat.Latent Heat: The heat required to convert a solid into a liquid or vapour (latent heat of fusion), or a liquid into a vapour(latent heat of evaporation), without change of temperature.Sensible Heat: The heat required to change the temperature of a body without changing its phase.So, basically as your question suggests, you need to evaluate the heat required to convert 2 kg of ice at 0 °C to water at 20 °C.. the total heat content will comprise of:Total heat = Heat required to convert 2 kg of ice to 2 kg of water at 0 °C + Heat required to convert 2 kg of water at 0 °C to 2 kg of water at 20 °C.[math]Heat = m hfg + m Cp ΔT[/math]Here, m ( mass of ice) = 2 kghfg (latent heat of fusion of ice) = 334 KJCp of water (specific heat) = 4.187 KJ/Kg-KΔT(Temperature difference) = 20 °CTherefore, Heat required = 2 x 334 + 2 x 4.187 x (20 - 0 )Heat reqd= 835.48 KJTherefore, to melt 2 kg of ice 835.48 KJ of heat is required.Happy Reading!
How much heat energy is required to raise the temperature of 8 kg water from 10℃ to 90℃?
The Old Thermodynamist says:Recall ΔH = m•Cp•ΔT where H is enthalpy (heat), m is mass Cp is heat capacity and ΔT is temperature change.The heat capacity of water is dependent on temperature, but is commonly taken as 4.186kJ/kg-KIn this case, we have 8kg of water and a temperature change ΔT = 80KSo ΔH = 8(4.186)(80) = 2679.0kJThe heat required is 2679.0kJ (which is 0.744kWh)