How much heat (energy) is required to convert 54.0 g (3.00 moles) of ice at -15.0°C to steam at a temperature of -105.0°C?
Details:
Specific heat capacity of ice: 2.05 Jg^-1 K^-1
Specific heat capacity of water: 4.18 Jg^-1 K^-1
Specific heat capacity of steam: 2.08 Jg^-1 K^-1
Molar heat of fusion for H2O: 6.02 kJ mol^-1
Molar heat of vaporization for H2O: 40.7 kJ mol^-1
Melting point of ice: 0.00°C
Boiling point of water: 373 K
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How much heat.....
Didn't we already play this game? This one just has more transitions: from ice at -15C to ice at 0C, melting of ice at 0C, from water at 0C to water at 100C, boiling at 100C, from steam at 100C to steam at 105C. You can solve the problem with one chain calculation
q = mcΔT(ice) + mHf + mcΔt(water) + mHv + mcΔT(steam) ............. factor out m
q = 54.0g(2.05 J/gC x 15C + 334 J/g + 4.18 J/gC x 100C + 2260 J/g + 2.02 J/gC x 5C)
q = 164853.9 J ............ round to three significant digits: 165,000 J ... or ... 165 kJ
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The first step is to determine the amount of heat energy that is required to increase the temperature of the ice from -15˚ to 0˚. The second step is to determine the amount of heat energy that is required to melt the ice. The third step is to determine the amount of heat energy that is required to increase the temperature of the water from 0˚C to 100˚C. The fourth step is to determine the amount of heat energy that is required to boil the water. The fifth step is to determine the amount of heat energy that is required to increase the temperature of the steam from 100˚ to 105˚
Q1 = 54 * 2.05 * 15 = 1660.5 J
Q2 = 3 * 6020 = 18,060 J
Q3 = 54 * 4.18 * 100 = 22,572 J
Q4 = 3 * 40,700 = 122,100 J
Q5 = 54 * 2.08 * 5 = 561.6 J
Total = 89,667 + 18,060 + 22,572 + 122,100 + 561.6 = 164,933.3 J
I hope this is helpful for you.
First you have to heat up the ice to 0 degrees Celsius.
Q = m * c * t
Q = amount of energy
m = mass
c = specific heat capacity
t = change in temperature (since a change of 1 degree kelvin and a change of 1 degree Celsius are the same, we can avoid converting too much)
Q[1] = 54 grams * 2.05 [J / (g * K)] * 15 K = 54 * 2.05 * 15 Joules = 810 * 2.05 J = 81 * 20.5 J = 1620 + 40.5 J = 1660.5 J
Now we need to melt that ice. So that requires the Molar Heat of Fusion
1 mole of water has a mass of 18 grams, so 54 grams = 3 moles
Q[2] = 3 moles * 6.02 kJ/mol = 3 * 6.02 kJ = 18.06 kJ = 18060 Joules
Now we need to heat this up to 100 degrees celsius
Q[3] = 54 grams * 4.18 J/(g * K) * 100 = 54 * 4.18 * 100 J = 54 * 418 J = 54 * (400 + 18) J = 21600 + 972 J = 22572 Joules
Now we have to vaporize it
Q[4] = 3 moles * 40.7 kJ/mol = 122.1 kJ = 122100 Joules
Now we need to raise it by 5 more degrees
Q[5] = 54 * 2.08 * 5 Joules = 27 * 2.08 * 10 J = 27 * 20.8 J = 540 + 27 - 5.4 J = 567 - 5.4 J = 561.6 J
Q[1] + Q[2] + Q[3] + Q[4] + Q[5] =>
1660.5 J + 18060 J + 22572 J + 122100 J + 561.6 J =>
(1660.5 + 561.6 + 18060 + 22572 + 122100) Joules =>
(2160.5 + 61.6 + 40572 + 60 + 122100) Joules =>
(2222.1 + 60 + 40000 + 572 + 122000 + 100) Joules =>
(2794.1 + 160 + 162000) Joules =>
(2954.1 + 162000) Joules =>
164954.1 Joules