How much energy is required to change a 40-g ice cube from ice at −5°C to steam at 140°C?
How much energy is required to change a 40-g ice cube from ice at −5°C to steam at 140°C?
Can someone please do out the problem? That would help a lot! Thanks!
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1. energy for heating 40g of ice from -5C to 0C = 40 x 0.5 x 5 = 100 cal.
2. heat for melting = 40 x 80 = 3200 cal.
3. heat for increasing the temperature of the water from 0 to 100 = 40 x 1 x 100 = 4000 cal.
4. heat of evaporation of 40g of water at 100C = 40 x 540 = 21600 cal.
5. heat for increasing the temperature of the steam from 100 to 140 = 40 x 1.8 x (140 - 100) = 2880 cal.
Adding up, 31780 cal.
For sp. heat of steam, see http://www.engineeringtoolbox.com/water-vapor-d_97...
Calculate the energy for heating 40g of ice from -5C to 0C, usung the specific heat of ice.
Next calculate the heat for melting, using the specific latent heat of ice to water.
Third, the heat for increasing the temperature of the water from 0 to 100, using the specific heat of water.
Fifth, the heat of evaporation of 40g of water at 100C.
Finally, the heat for increasing the temperature of the steam from 100 to 140, using the specific heat of steam.
Finally add all these energies.
to develop the temperature to 0°C, the warmth required is Q1 = mc?T the place m = fifty 5/1000 kg c = 1943 J/kg.°C the place we take the c cost at -20°C (see the ref under) ?T = 0 - (-40) = 40°C Q1 = 4274.6 J to transform 0.055kg of water at 0°C to ice at 0°C, a factor replace happens so we use Q2 = mL the place L = 3.33 * 10^5 J/kg is the latent warmth of fusion for water Q2 = 18315 J to develop the temperature to one hundred°C from 0°C, the warmth required is Q3 = mc?T the place c = 4182 J/kg.°C for water at 50°C would be used (see the ref under) ?T = one hundred - 0 = one hundred°C So Q3 = 23001 J to transform water at one hundred°C to steam at one hundred°C, the warmth required is this autumn = mL the place L = 2279 kJ/kg is the latent warmth of evaporation this autumn = 125345 J to develop the temperature to one hundred forty°C from vapour at one hundred°C, the warmth required is Q5 = mc?T the place c = 4248 J/kg°C for steam at one hundred twenty°C (yet another ref provides c = specific warmth skill water vapor at one hundred twenty°C - a million.996 kJ/kgC on the boiling element. in all danger has to do with the rigidity which grow to be no longer specfied) and ?T = one hundred forty - one hundred = 40 Q5 = 9345.6 J (Q5' = 4391.2 J if we use the 2d cost for c) the warmth required is the sum Q Q = Q1 + Q2 + Q3 + this autumn + Q5 (Q5') = 180281.2 J (175326.8 J) = a million.803 * 10^5 J (a million.753 * 10^5 J). desire this helps