{"id":1681,"date":"2018-04-03T11:45:34","date_gmt":"2018-04-03T03:45:34","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1681"},"modified":"2023-08-10T00:22:19","modified_gmt":"2023-08-10T00:22:19","slug":"heat-and-change-of-phase-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/heat-and-change-of-phase-problems-and-solutions.htm","title":{"rendered":"Heat and change of phase \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Heat and change of phase \u2013 problems and solutions<\/span><\/p>\n

1. Based on the graph, what is the heat<\/a> absorbed by 5-kg water during process C-D ? Specific heat<\/a> of water = 4,200 J\/kg o<\/sup>C.<\/span><\/p>\n

Known :\"Heat<\/u><\/span><\/p>\n

Mass<\/a> (m) = 5 kg<\/span><\/p>\n

Specific heat of water (c) = 4200 J\/kg o<\/sup>C <\/span><\/p>\n

Initial temperature (T1<\/sub>) = 0 o<\/sup>C<\/span><\/p>\n

Final temperature (T2<\/sub>) = 10 o<\/sup>C<\/span><\/p>\n

The change in temperature (\u0394T) = 10 o<\/sup>C – 0 o<\/sup>C = 10 o<\/sup>C <\/span><\/p>\n

Wanted :<\/u> Heat (Q)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Q = m c \u0394T = (5 kg)(4200 J\/kg o<\/sup>C)(10 o<\/sup>C) = (5)(4200 J)(10) = (50)(4200 J) = 210,000 Joule<\/span><\/p>\n

2. What is the heat absorbed by 20 gram ice during process A-C, if the specific heat of ice is 2100 J\/kg o<\/sup>C and heat of fusion for ice is 336,000 J\/kg.<\/span><\/p>\n

\"HeatKnown :<\/u><\/span><\/p>\n

Mass of ice (m) = 200 gram = 200\/1000 kilogram = 0.2 kilogram<\/span><\/p>\n

Specific heat (c) = 2100 J\/kg o<\/sup>C<\/span><\/p>\n

Heat of fusion for ice (LF<\/sub>) = 336,000 J\/kg <\/span><\/p>\n

Wanted:<\/u> Heat absorbed by ice during process A-C<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

During process A-B, heat is absorbed by ice to change its temperature from -10o<\/sup>C to 0o<\/sup>C.<\/span><\/p>\n

Q = m c \u0394T<\/span><\/p>\n

Q = heat, m = mass of ice, c = specific heat of ice, \u0394T = the change in temperature<\/i><\/span><\/p>\n

Q1 <\/sub>= m c \u0394T = (0.2)(2100)(0-(-10) = (0.2)(2100)(10) = (2)(2100) = 4200 Joule <\/span><\/p>\n

During process B-C, heat is absorbed to melting all ice become water. In this process only changes in phase and there is no change in temperature. The ice and water temperature remains 0o<\/sup>C.<\/span><\/span><\/p>\n

Heat absorbed during process B-C :<\/span><\/p>\n

Q2 <\/sub>= m LF<\/sub> = (0.2 kg)(336,000 J\/kg) = 67,200 Joule<\/span><\/p>\n

Heat absorbed during process A-C :<\/span><\/p>\n

Q = Q1 <\/sub>+ Q2 <\/sub>= 4200 Joule + 67,200 Joule = 71,400 Joule <\/span><\/p>\n

3. A 2-kg water is heated, and heat of vaporization for water = 2.27 x 106<\/sup> J\/kg, specific heat for water = 4,200 J\/kg o<\/sup>C and atmosphere pressure is 1 atm, what is the heat absorbed by water during process B-C ?<\/span><\/p>\n

Known :\"Heat<\/u><\/span><\/p>\n

Mass (m) = 2 kg<\/span><\/p>\n

Heat of vaporization for water (LV<\/sub>) = 2.27 x 106<\/sup> J\/kg <\/span><\/p>\n

Specific heat for water (c) = 4,200 J\/kg o<\/sup>C <\/span><\/p>\n

Wanted :<\/u> Heat absorbed during process B-C.<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

In process A-B, heat absorbed to increases the temperature of water from 60o<\/sup>C to 100o<\/sup>C. <\/span><\/p>\n

Q = m LV<\/sub> = (2 kg)(2.27 x 106<\/sup> J\/kg) = 4.54 x 106<\/sup> Joule = 4540 x 103 <\/sup>Joule = 4540 kilo Joule <\/span><\/p>\n

4. Based on figure below, if mass of ice is 500 gram, specific heat is 2100 J\/kg o<\/sup>C and latent heat of fusion of ice is 336,000 J\/kg. Determine heat required in process P\u2013Q\u2013R.<\/span><\/p>\n

Known :\"Heat<\/u><\/span><\/p>\n

Mass of ice (m) = 500 gram = 500\/1000 kg = 0.5 kg <\/span><\/p>\n

Specific heat for ice (c) = 2100 J\/kg o<\/sup>C<\/span><\/p>\n

Latent heat<\/a> of fusion (LF<\/sub>) = 336,000 J\/kg <\/span><\/p>\n

Wanted :<\/u> Heat<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

P<\/span>rocess<\/span> P-Q = heat absorbed to raise the ice temperature from -4o<\/sup>C to 0o<\/sup>C<\/span><\/p>\n

P<\/span>rocess <\/span>Q-R = heat absorbed by ice to melting all the ice into water<\/span><\/p>\n

Heat absorbed during process P-Q calculated using equation :<\/span><\/p>\n

Q = m c \u0394T<\/span><\/p>\n

Q = heat, m = mass, c = specific heat of ice, \u0394T = the change in temperature<\/i><\/span><\/p>\n

Heat absorbed ice during process P-Q calculated using this equation :<\/span><\/p>\n

Q = (0.5 kg)(2100 J\/kg o<\/sup>C)(0o<\/sup>C-(-4o<\/sup>C))<\/span><\/p>\n

Q = (0.5 kg)(2100 J\/kg o<\/sup>C)(0o<\/sup>C + 4o<\/sup>C)<\/span><\/p>\n

Q = (0.5 kg)(2100 J\/kg o<\/sup>C)(4o<\/sup>C)<\/span><\/p>\n

Q = (0.5)(2100 J)(4)<\/span><\/p>\n

Q = 4200 Joule<\/span><\/p>\n

Heat absorbed during process Q-R calculated using this equation :<\/span><\/p>\n

Q = m LF<\/sub> <\/span><\/p>\n

Q = (0.5 kg)(336,000 J\/kg)<\/span><\/p>\n

Q = (0.5)(336,000 J)<\/span><\/p>\n

Q = 168,000 Joule<\/span><\/p>\n

Total heat :<\/span><\/p>\n

Q = 4200 Joule + 168,000 Joule <\/span><\/p>\n

Q = 172,200 Joule<\/span><\/p>\n

Q = 172.2 kilo Joule<\/span><\/p>\n

Q = 172.2 kJ<\/span><\/p>\n

5. Ice with mass of 50 gram heated from -5o<\/sup>C into water at 60o<\/sup>C. If heat of fusion for ice is 80 cal\/gram, specific heat for ice is 0.5 cal\/gram o<\/sup>C, specific heat for water is 1 cal\/gram o<\/sup>C, determine heat required in process C-D.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of ice = mass of water (m) = 50 gram \"Heat<\/span><\/p>\n

Latent heat of fusion for ice (LF<\/sub>) = 80 cals\/gram<\/span><\/p>\n

Specific heat for ice (c ice) = 0.5 cal\/gram o<\/sup>C<\/span><\/p>\n

Specific heat for water (c water) = 1 cal\/gram o<\/sup>C<\/span><\/p>\n

Wanted:<\/u> Heat required in process C-D<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

In process A to B, heat used to raise the temperature of ice from -5o<\/sup>C to 0o<\/sup>C. Heat calculated using this equation : Q = m c \u0394T, where Q = heat, m = mass of ice, c = specific heat for ice, \u0394T = the change in temperature from -5o<\/sup>C to 0o<\/sup>C.<\/span><\/p>\n

In process B to C, heat used to melting all ice to water. In this process, temperature not changed. Heat calculated using this equation: Q = m LF<\/sub>, where Q = heat, m = mass of ice, LF <\/sub>= latent heat of fusion of ice.<\/span><\/p>\n

In process C to D, heat used to raise the temperature of water from 0o<\/sup>C to 60o<\/sup>C. Heat calculated using this equation: Q = m c \u0394T, where Q = heat, m = mass of water, c = specific heat of water, \u0394T = the change in temperature from 0o<\/sup>C to 60o<\/sup>C.<\/span><\/p>\n

The heat required from process C to D :<\/span><\/p>\n

Q = m c \u0394T = (50 gram)(1 cal\/gram o<\/sup>C)(60o<\/sup>C-0o<\/sup>C) = (50)(1)(60) cal = 3000 cal<\/span><\/p>\n

6. If the mass of ice is 1 kg, the specific heat of ice is 2100 J\/kgo<\/sup>C, latent heat of fusion for ice is 336,000 J\/kg and specific heat of water is 4200 J\/kgo<\/sup>C, then determine heat required in process P-Q-R.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of ice = 1 kg\"Heat<\/span><\/p>\n

Latent heat of fusion of ice (LF<\/sub>) = 336,000 J\/kg<\/span><\/p>\n

Specific heat of ice (c ice) = 2,100 J\/kgo<\/sup>C<\/span><\/p>\n

Specific heat of water (c water) = 4,200 J\/kgo<\/sup>C<\/span><\/p>\n

Wanted:<\/u> Heat required in process P-Q-R <\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

In process P to Q, heat used to raise the temperature of ice from -5o<\/sup>C to 0o<\/sup>C. Heat calculated using this equation : Q = m c \u0394T, where Q = <\/i>heat<\/i>, m = mass <\/i>of ice<\/i>, c = <\/i>specific heat of ice, <\/i>\u0394T = <\/i>the change in temperature from <\/i>-5<\/i>o<\/i><\/sup>C <\/i>to <\/i>0<\/i>o<\/i><\/sup>C.<\/i><\/span><\/p>\n

Heat required in process P to Q :<\/span><\/p>\n

Q = m c \u0394T = (1)(2100)(0-(-5)) = (2100)(5) = 10500 Joule <\/span><\/p>\n

In process Q to R, the heat required to melting all ice into water. In this process, temperature is unchanged. Heat calculated using this equation: Q = m L<\/i>F<\/i><\/sub>, <\/i>where <\/i>Q = <\/i>heat<\/i>, m = mass <\/i>of ice<\/i>, L<\/i>F <\/i><\/sub>= <\/i>latent heat of fusion of ice<\/i>.<\/i><\/span><\/p>\n

Heat required in process Q to R :<\/span><\/p>\n

Q = m LF <\/sub>= (1)(336,000) = 336,000 Joule<\/span><\/p>\n

Total heat :<\/span><\/p>\n

10,500 + 336,000 = 346,500 Joule<\/span><\/p>\n

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  1. What is latent heat?<\/strong><\/span>\n