Compute the thermal energy that is necessary to convert 5.00 kg of ice at -20.0

Question

Compute the thermal energy that is necessary to convert 5.00 kg of ice at -20.0oC to superheated steam at 120oC. Show ALL Work.

Information from Lab Conducted:

Heat (Q) can be defined as the transfer of energy due to temperature difference. When two bodies having different temperatures are brought in contact with each other, heat energy will flow from the hotter body to the cooler body till both of them reach the same temperature. Then the system is said to be in equilibrium Since heat is defined as a form of energy, its unit in SI isJ. The Specific heat c is defined as the heat required to raise the temperature of 1.00kg of mass(m) by 1.00 oC. The latent heat of fusion is the amount of heat needed per kilogram to melt ice. Calorimetry is the measurement of heat and is performed in a device called calorimeter. Calorimeter works on the basic principle of conservation of energy The thermal energy lost by a body should be gained by those surrounding the body Thermal energy lost = Thermal energy gained Experiment steps Mass Step-1: Measure the mass of the calorimeter along with the stirrer (m(cal-stirrer)). Step-2: Fill approximately half the calorimeter cup with water and measure the mass (m(caleStirrer+Water). The difference between m(Cal-stirrer*Water) and miCal+stirrer) will give the mass of water - mw. Record all the masses in the observations table Temperature Step-3: Stir the water in the calorimeter cup using the stirrer. Use the insulating cap on the stirrer so that the system is well insulated. Make sure that the thermometer reading is not changing and record the value of initial temperature (Ti) Step-5: Put the solid ice into the water (Dry it with a paper towel so that the water around the ice is removed.) and close the calorimeter. Keep stirring till the ice melts completely. Keep stirring till an equilibrium temperature is reached. Measure and record the final temperature (T)

Explanation / Answer

given m = 5 kg ice
Ti = -20 C
Tf = 120 C

now, heat capacity of ice, Ci = 2108 J / kg K
heat capacity of water = Cw = 4187 J / kg K
heat capacity of steam = Cs = 1996 J / kg K

Latent heat of fusion = Lf = 334,000 J/ kg
Latent heat of vapourisation, Lv = 2,230,000 J/kg

hence total heat required = m(Ci(0 - Ti) + Lf + Cw*100 + Lv + Cs*(Tf - 100))
Q = 15323900 J = 15.3239*10^6 J = 15.3239 MJ

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