Question

A group of students place a 0.045 kg water sample, initial temperature 21.5 degrees Celsius, on a heat source. The water is then warmed up to its boiling point then further heated until it is completely evaporated. Note: Both the heat source and the water are in a thermally isolated container.

i. Find the total heat absorbed by the water during this process.

ii. A similar experiment is performed on an unknown liquid, mass=0.055kg with initial temperature 21.5 degrees Celsius. During the experiment, the unknown liquid absorbs a total of 2.03x10^4 J to completely evaporate the liquid. Note: The specific heat for the unknown liquid is 1130J/kg*K and its boiling point temperature is 110.5 degrees Celsius. Determine the latent heat of vaporization for this liquid.

Answer 1

Solution :-

Solution to this problem is based on the concepts of calorimetry, part (i) and (ii) are based on same formula where total heat absorbed by water/unkown liquid = heat required to rasie temperature from initial temperature(21.5) to boiling point (100⁰C for water & 110.5⁰C for unknown liquid) + heat required to change phase of liquid at constant temperature (constant temperature being boiling point for liquid). I solved part (i) in cgs units and finally converted answer into SI units where part (ii) is solved entirely in SI units as every data for part (ii) is provided in SI units. dT or deltaT is same in ⁰C or Kelvin.

Please read solution carefully and feel free to ask any doubt you face while understanding my solution. I would be happy to help.

0.045 X 100 Salu & mass of water (ma) = 0.045 kg nSg. initial temperature of Water (Twi) = 21.5°C Heat absorbed by water that required to raise temp. of 0.045 kg of water from 21.5 c to look & tent heat required for phase change. Latent heat - Heat required for phase change of unit mass. specific heat of water (a I callg-'c - I callgok Heat required to raise temp. of water @ = medt to look = (0.045 x 1000g) XIX (100 - 21.5) Q = 45X14 78.5 - Q = 3532.5 cal katent heat at 100 c (Open charge Heat required for phase Change from water at tool to vapours & mdr) where ty is latent heat of vapourization do= 540 Callg 10.045x100 og/x 540 249300 cal Heat absorbed by weater = 3532.5 + 24 300 O phan change = 2 18 32.5 cal Ical = 4.2 Joules - 21832.5 x 4.2 Joules =116896.5 Joule

10 110.5°C mass of unknown liquid (m) = 0.055 kg = 55g initial temperature (T;) = 21.5°C (0) Heat absorbed by liquid to completely evaporate = 2-03 xlo"J (Q (heat absorbed by liquid to completely and evapolate Heat required to raise temp from 21.50 totsen & Heat required for phase change. specific heat (C) of unknown liquid = 1130J/kg k Heat required to raise temperature from 21.50 to 110.5c of 0.055 kg of unknown liquid her dt or at is temperature differenca same for oC or Kelvin (K) & 80 DT = (Tf - T: ) DT = (Tp-T.) = (110.5c-21.5c) = 89.c. (in "C) OT = (TP-T ) = (383.5k-294.5k) = 89 K (in Kelvin) from equation Heat required to raise temperature from 21.5°C (294.5k) to 110.5°c (383.5x) of 0.055 kg of unkown liquid= Դ Դ Դ Դ Դ Դ Դ Դ Դ Դ Դ Դ Ր Ր Ր Ո Ր mcdT and is howe gwe 0.055 X 1130 X 89 Q = 5531.35 Joules Total Heat absorbed = 5531.35 + Heat absorbed for phase change 2.03 x 10" = 5531.35 + Heat absorbed for phase change. heat absorbed for 2.03x10h - 5531.35 phase change heat absorbed for - 14768.65 phase change 9

where & is latent heat of Vapourization of unknown liquid Heat aboooked for phase change = M L from equation ③ and (3) I we have ML = 14768.65 0.055XL = 14768.65 L = 14768-65 0.055 268520.9 J/kg