{"id":2104,"date":"2018-04-24T10:13:31","date_gmt":"2018-04-24T02:13:31","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2104"},"modified":"2023-08-09T04:03:28","modified_gmt":"2023-08-09T04:03:28","slug":"temperature-and-heat-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/temperature-and-heat-problems-and-solutions.htm","title":{"rendered":"Temperature and heat \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Temperature and heat \u2013 problems and solutions<\/p>\n

1. On a thermometer X, the freezing point of water at -30o<\/sup> and the boiling point of water at 90o<\/sup>. 60O<\/sup>X = \u2026.. o<\/sup>C.<\/span><\/span><\/p>\n

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

The freezing point of water = -30o<\/sup> <\/span><\/span><\/p>\n

The boiling point of water = 90o<\/sup> <\/span><\/span><\/p>\n

Wanted :<\/u> 60o<\/sup>X = ….. o<\/sup>C<\/span><\/span><\/p>\n

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

On the Fahrenheit scale, the freezing point of water is 32o<\/sup>F and the boiling point of water is 212o<\/sup>F. Between the freezing point and the boiling point, 212o<\/sup> \u2013 32o<\/sup> = 180o<\/sup>. <\/span><\/span><\/p>\n

On the Celsius scale, the freezing point of water is 0o<\/sup>C and the boiling point of water is 100o<\/sup>C. Between the freezing point and the boiling point, 100o<\/sup> \u2013 0o<\/sup> = 100o<\/sup>. <\/span><\/span><\/p>\n

On the X scale, the freezing point of water is -30o<\/sup>X and the boiling point of water is 90o<\/sup>X. Between the freezing point and the boiling point, 90o<\/sup> – (-30o<\/sup>) = 90o<\/sup> + 30o <\/sup>= 120o<\/sup>. <\/span><\/span><\/p>\n

Change the X scale to the Celsius scale :<\/span><\/span><\/p>\n

\"Temperature<\/p>\n

Thermal expansion<\/b><\/span><\/span><\/a><\/p>\n

2<\/span><\/span>. <\/span><\/span>A metal rod heated from 30<\/span><\/span>o<\/span><\/span><\/sup>C to <\/span><\/span>80<\/span><\/span>o<\/span><\/span><\/sup>C. <\/span><\/span>The final length of the rod is <\/span><\/span>115 cm. <\/span><\/span>The coefficient of linear expansion<\/a> is <\/span><\/span>3.10<\/span><\/span>-3<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>. <\/span><\/span>What is the initial length of the metal rod?<\/span><\/span><\/p>\n

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

The initial temperature <\/span><\/span>(T<\/span><\/span>1<\/span><\/span><\/sub>) = 30<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

The final temperature <\/span><\/span>(T<\/span><\/span>2<\/span><\/span><\/sub>) = 80<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

The change in temperature <\/span><\/span>(\u0394T) = 80<\/span><\/span>o<\/span><\/span><\/sup>C – 30<\/span><\/span>o<\/span><\/span><\/sup>C = 50<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

The coefficient of linear expansion <\/span><\/span>(\u03b1) = 3.10<\/span><\/span>-3<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

The final length of the metal (L) = 115 cm<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The initial length of the metal rod <\/span><\/span>(L<\/span><\/span>o<\/span><\/sub>)<\/span><\/span><\/p>\n

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

The equation of the linear expansion :<\/span><\/span><\/p>\n

L = L<\/span><\/span>o<\/span><\/sub> + \u0394L<\/span><\/span><\/p>\n

L = L<\/span><\/span>o<\/span><\/sub> + \u03b1 L<\/span><\/span>o<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

L = L<\/span><\/span>o<\/span><\/sub> (1 + \u03b1 \u0394T)<\/span><\/span><\/p>\n

115 = L<\/span><\/span>o<\/span><\/sub> (1 + 3.10<\/span><\/span>-3<\/span><\/span><\/sup>.50) <\/span><\/span><\/p>\n

115 = L<\/span><\/span>o<\/span><\/sub> (1 + 150.10<\/span><\/span>-3<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

115 = L<\/span><\/span>o<\/span><\/sub> (1 + 0.15)<\/span><\/span><\/p>\n

115 = L<\/span><\/span>o<\/span><\/sub> (1.15)<\/span><\/span><\/p>\n

L<\/span><\/span>o<\/span><\/sub> = 115 \/ 1.15<\/span><\/span><\/p>\n

L<\/span><\/span>o<\/span><\/sub> = 100 cm <\/span><\/span><\/p>\n

3<\/span><\/span>. <\/span><\/span>The initial length of a brass rod is <\/span><\/span>40 cm. <\/span><\/span>After heated, the final length of the brass is 40.04 cm and the final temperature is <\/span><\/span>80<\/span><\/span>o<\/span><\/span><\/sup>C. <\/span><\/span>If the coefficient of linear expansion of the brass is <\/span><\/span>2.0 x 10<\/span><\/span>-5<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>–<\/span><\/span><\/sup>1, <\/span><\/span>what is the initial temperature of the brass rod.<\/span><\/span><\/p>\n

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

The final temperature <\/span><\/span>(T<\/span><\/span>2<\/span><\/sub>) = 80<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

The initial length <\/span><\/span>(L<\/span><\/span>o<\/span><\/sub>) = 40 cm<\/span><\/span><\/p>\n

The final length (L) = 40.04 cm <\/span><\/span><\/p>\n

The increase in length (\u0394L) = 40.04 cm \u2013 40 cm = 0.04 cm<\/span><\/span><\/p>\n

The coefficient of linear expansion <\/span><\/span>(\u03b1) = 2.0 x 10<\/span><\/span>-5<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

Solution :<\/u><\/span><\/span> Initial temperature <\/span><\/span>(T<\/span><\/span>1<\/span><\/sub>)<\/span><\/span><\/p>\n

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

The equation of the linear expansion :<\/span><\/span><\/p>\n

L = L<\/span><\/span>o<\/span><\/sub> + \u03b1 L<\/span><\/span>o<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

L – L<\/span><\/span>o<\/span><\/sub> = \u03b1 L<\/span><\/span>o<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

\u0394L = \u03b1 L<\/span><\/span>o<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

\u0394L = \u03b1 L<\/span><\/span>o<\/span><\/sub> (T<\/span><\/span>2<\/span><\/sub> \u2013 T<\/span><\/span>1<\/span><\/sub>) <\/span><\/span><\/p>\n

0.04 = (2.0 x 10<\/span><\/span>-5<\/span><\/span><\/sup>)(40)(80 \u2013 T<\/span><\/span>1<\/span><\/sub>)<\/span><\/span><\/p>\n

0..04 = (80 x 10<\/span><\/span>-5<\/span><\/span><\/sup>)(80 – T<\/span><\/span>1<\/span><\/sub>)<\/span><\/span><\/p>\n

0.04 = 0.0008 (80 \u2013 T<\/span><\/span>1<\/span><\/sub>)<\/span><\/span><\/p>\n

0.04 = 0.064 \u2013 0.0008 T<\/span><\/span>1 <\/span><\/sub><\/p>\n

0..0008 T<\/span><\/span>1<\/span><\/sub> = 0.064 \u2013 0.040<\/span><\/span><\/p>\n

0.0008 T<\/span><\/span>1<\/span><\/sub> = 0.024<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> = 30<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

Heat transfer conduction<\/b><\/span><\/span><\/a><\/p>\n

4. Two metal rods with the same size but different type, as shown in figure below. The thermal conductivity of metal I = 4 times the thermal conductivity of metal II. What is the temperature between both metals. <\/span><\/span><\/p>\n

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

The size of both rods is the same.\"Temperature<\/span><\/span><\/p>\n

The thermal conductivity of metal I = 4k <\/span><\/span><\/p>\n

The thermal conductivity of metal II = k <\/span><\/span><\/p>\n

The temperature of the one end of metal I = 500<\/sup> C <\/span><\/span><\/p>\n

The temperature of the one end of metal II = 00<\/sup> C <\/span><\/span><\/p>\n

Wanted:<\/u> The temperature between both the metal rods<\/span><\/span><\/p>\n

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

The equation of the heat conduction :<\/span><\/span><\/p>\n

\"Temperature<\/p>\n

Q\/t = <\/i>the rate of heat conduction<\/i>, k = <\/i>thermal conductivity<\/i>, A = <\/i>the cross-sectional area<\/i>, T<\/i>1<\/i><\/sub>-T<\/i>2<\/i><\/sub> = <\/i>the change in temperature<\/i>, l = <\/i>the length of the rod.<\/i><\/span><\/span><\/p>\n

The temperature between both rods :<\/span><\/span><\/p>\n

\"Temperature<\/p>\n

The temperature at the center between both the metals rods is 40o<\/sup>C.<\/span><\/span><\/p>\n

5. <\/span><\/span><\/p>\n

(1) Conductivity of metal<\/span><\/span><\/p>\n

(2) The difference of temperature<\/span><\/span><\/p>\n

(3) The length of metal<\/span><\/span><\/p>\n

(4) Mass of metal<\/span><\/span><\/p>\n

Factors that determine the rate of heat conduction on metals are…..<\/span><\/span><\/p>\n

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

Based on the equation of heat conduction, factors that determine the rate of heat conduction on metals are conductivity of metal (k), the difference of temperature (T) and the length of metal (l). <\/span><\/span><\/p>\n

6. Two rods of the same size but different type, as shown in the figure below. The thermal conductivity of rod P is 2 times the thermal conductivity of rod Q. What is the temperature between both rods.<\/span><\/span><\/p>\n

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

Both rods have the same size.\"Temperature<\/span><\/span><\/p>\n

Thermal conductivity of rod P (kP<\/sub>) = 2k<\/span><\/span><\/p>\n

The thermal conductivity of rod Q (kQ<\/sub>) = k<\/span><\/span><\/p>\n

Wanted:<\/u> The temperature between both rods<\/span><\/span><\/p>\n

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

The equation of the heat conduction :<\/span><\/p>\n

\"Temperature<\/p>\n

Q\/t = <\/i>the rate of heat conduction<\/i>, k = <\/i>thermal conductivity<\/i>, A = <\/i>the cross-sectional area<\/i>, T<\/i>1<\/i><\/sub>-T<\/i>2<\/i><\/sub> = <\/i>the change in temperature<\/i>, l = <\/i>the length of the rod.<\/i><\/span><\/span><\/p>\n

The temperature at the center between both rods :<\/span><\/span><\/p>\n

\"Temperature<\/p>\n

Black principle<\/b><\/span><\/span><\/a><\/p>\n

8. 100-gram oil at 20o<\/sup>C and 50-gram iron at 75 o<\/sup>C are placed in 200-gram iron container. The increase in temperature of the container is 5o<\/sup>C and the specific heat of oil is 0.43 cal\/g o<\/sup>C. What is the specific heat of the iron?<\/span><\/span><\/p>\n

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

Mass of iron container (m) = 200 gr<\/span><\/span><\/p>\n

The initial temperature of the iron container (T1<\/sub>) = the temperature of oil = 20o<\/sup>C <\/span><\/span><\/p>\n

The final temperature of the iron container (T2<\/sub>) = 20o<\/sup>C + 5o<\/sup>C = 25o<\/sup>C<\/span><\/span><\/p>\n

Mass of oil (m) = 100 gram<\/span><\/span><\/p>\n

The specific heat of oil (coil<\/sub>) = 0.43 cal\/g o<\/sup>C <\/span><\/span><\/p>\n

The initial temperature of oil (T1<\/sub>) = 20o<\/sup>C<\/span><\/span><\/p>\n

The final temperature of oil (T2<\/sub>) = 20o<\/sup>C + 5o<\/sup>C = 25o<\/sup>C<\/span><\/span><\/p>\n

Mass of iron (m) = 50 gram<\/span><\/span><\/p>\n

The initial temperature of oil (T1<\/sub>) = 75o<\/sup>C<\/span><\/span><\/p>\n

The final temperature of oil (T2<\/sub>) = 25o<\/sup>C<\/span><\/span><\/p>\n

Wanted :<\/u> The specific heat of iron (c iron)<\/span><\/span><\/p>\n

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

Heat released by iron :<\/span><\/span><\/p>\n

Q = m c \u0394T = (50)(c)(75-25) = (50)(c)(50) = 2500c calorie<\/span><\/span><\/p>\n

Heat absorbed by the iron container :<\/span><\/span><\/p>\n

Q = m c \u0394T = (200)(c)(25-20) = (200)(c)(5) = 1000c calorie<\/span><\/span><\/p>\n

Heat absorbed by oil :<\/span><\/span><\/p>\n

Q = m c \u0394T = (100)(0.43)(25-20) = (43)(5) = 215 calorie<\/span><\/span><\/p>\n

Black principle states that in a isolated system, heat released by the hotter object, absorbed by the cooler object. <\/span><\/span><\/p>\n

Q release = Q absorb<\/span><\/span><\/p>\n

2500c = 1000c + 215<\/span><\/span><\/p>\n

2500c – 1000c = 215<\/span><\/span><\/p>\n

1500c = 215<\/span><\/span><\/p>\n

c = 215\/1500<\/span><\/span><\/p>\n

c = 0.143 cal\/g o<\/sup>C <\/span><\/span><\/p>\n

9. A 200-gram water at 20\u00b0C placed in 50-gram ice at -2\u00b0C. If the change of heat just between water and ice, what is the final temperature of the mixture? The specific heat of water is 1 cal\/gr\u00b0C, the specific heat of ice is 0.5 cal\/gr\u00b0C, the heat of fusion for ice is 80 cal\/gr.<\/span><\/span><\/p>\n

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

Mass of water (mwater<\/sub>) = 200 gram<\/span><\/span><\/p>\n

Temperature of water (Twater<\/sub>) = 20o<\/sup>C<\/span><\/span><\/p>\n

The specific heat of water (cwater<\/sub>) = 1 cal\/gr\u00b0C<\/span><\/span><\/p>\n

Mass of ice (mice<\/sub>) = 50 gram<\/span><\/span><\/p>\n

Temperature of ice (Tice<\/sub>) = -2o<\/sup>C<\/span><\/span><\/p>\n

The specific heat of ice (cice<\/sub>) = 0.5 cal\/gr\u00b0C <\/span><\/span><\/p>\n

The heat of fusion for ice (L) = 80 cal\/gr <\/span><\/span><\/p>\n

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

Heat to increases ice from -2o<\/sup>C to 0o<\/sup>C :<\/span><\/span><\/p>\n

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

Q = (50 gram)(0.5 cal\/gr\u00b0C)(0o<\/sup>C – (-2o<\/sup>C))<\/span><\/span><\/p>\n

Q = (50)(0.5 cal)(2)<\/span><\/span><\/p>\n

Q = 50 calorie<\/span><\/span><\/p>\n

Heat for melting all ice :<\/span><\/span><\/p>\n

Q = m L = (50 gram)(80 cal\/gram) = 4000 calorie<\/span><\/span><\/p>\n

Heat for decrease temperature of all water from 20o<\/sup>C to 0o<\/sup>C :<\/span><\/span><\/p>\n

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

Q = (200 gram)(1 cal\/gr\u00b0C)(0o<\/sup>C \u2013 (20o<\/sup>C))<\/span><\/span><\/p>\n

Q = (200)(1 cal)(-20)<\/span><\/span><\/p>\n

Q = -4000 calorie<\/span><\/span><\/p>\n

Plus sign indicates that the heat is added, the minus sign indicates that heat released.<\/i><\/span><\/span><\/p>\n

50-calorie of heat needed to increase the temperature of ice to 0o<\/sup>C and 4000-calorie needed to melting all ice. Total heat = 4050 calorie. The heat released by water is 4000 calorie.<\/span><\/span><\/p>\n

Some of the ice not melting, so the final temperature of ice and water is 0o<\/sup>C.<\/span><\/span><\/p>\n

10. A 200-gram aluminum at 20o<\/sup>C placed in 100-gram water at 80o<\/sup>C in a container. The specific heat of\u00a0aluminum is 0.22 cal\/g o<\/sup>C and the specific heat of water is 1 cal\/g o<\/sup>C. What is the final temperature of aluminum?<\/span><\/span><\/p>\n

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

Mass of aluminum = 200 gram<\/span><\/span><\/p>\n

Temperature of aluminum = 20o<\/sup>C<\/span><\/span><\/p>\n

Mass of water = 100 gram<\/span><\/span><\/p>\n

Temperature of water = 80o<\/sup>C<\/span><\/span><\/p>\n

The specific heat of aluminum = 0.22 cal\/g o<\/sup>C <\/span><\/span><\/p>\n

The specific heat of water = 1 cal\/g o<\/sup>C<\/span><\/span><\/p>\n

Wanted:<\/u> The final temperature of aluminum
\n<\/span><\/span><\/p>\n

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

Aluminum and water in thermal equilibrium so that the final temperature of aluminum = the final temperature of water.<\/span><\/span><\/p>\n

Heat released by hot water (Q release) = heat absorbed by aluminum (Q absorb)<\/u><\/span><\/span><\/p>\n

mwater<\/sub> c (\u0394<\/span>T) = maluminum<\/sub> c (\u0394<\/span>T) <\/span><\/span><\/p>\n

(100)(1)(80 – T) = (200)(0.22)(T \u2013 20)<\/span><\/span><\/p>\n

(100)(80 – T) = (44)(T \u2013 20)<\/span><\/span><\/p>\n

8000 \u2013 100T = 44T \u2013 880<\/span><\/span><\/p>\n

8000 + 880 = 44T + 100T<\/span><\/span><\/p>\n

8880 = 144T<\/span><\/span><\/p>\n

T = 62o<\/sup>C<\/span><\/span><\/p>\n

11. A 50-gram metal at 85 \u00b0C placed in 50 gram water at 29.8 \u00b0C. The specific heat of water = 1 cal.g \u20141<\/sup> .\u00b0C\u20141<\/sup>. The final temperature is 37 \u00b0C. What is the specific heat of metal.<\/span><\/span><\/p>\n

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

Mass of metal (mmetal<\/sub>) = 50 gram<\/span><\/span><\/p>\n

Temperature of metal = 85o<\/sup>C <\/span><\/span><\/p>\n

Mass of water (mwater<\/sub>) = 50 gram<\/span><\/span><\/p>\n

Temperature of water = 29,8o<\/sup>C<\/span><\/span><\/p>\n

The specific heat of water (cwater<\/sub>) = 1 cal.g -1<\/sup> .\u00b0C-1<\/sup><\/span><\/span><\/p>\n

The final temperature of water = 37o<\/sup>C<\/span><\/span><\/p>\n

Wanted :<\/u> The specific heat of metal (c metal)<\/span><\/span><\/p>\n

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

Heat released by hot metal (Q release) = heat absorbed by water (Q absorb)<\/u><\/span><\/span><\/p>\n

mmetal<\/sub> c (\u0394<\/span>T) = mwater<\/sub> c (\u0394<\/span>T) <\/span><\/span><\/p>\n

(50)(c)(85 – 37) = (50)(1)(37 \u2013 29.8)<\/span><\/span><\/p>\n

(c)(85 \u2013 37) = (1)(37 \u2013 29.8)<\/span><\/span><\/p>\n

48 c = 7.2<\/span><\/span><\/p>\n

c = 0.15 cal.g -1<\/sup> .\u00b0C-1<\/sup><\/span><\/span><\/p>\n

12. A block of ice with mass of 50-gram at 0\u00b0C and 200-gram water at 30\u00b0C, placed in a container. . If the specific heat of water is 1 cal.g– 1<\/sup> \u00b0C \u20131<\/sup> and the heat of fusion for ice is 80 cal.g \u2013<\/sup>1. What is the final temperature of the mixture.<\/span><\/span><\/p>\n

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

Mass of ice (mice<\/sub>) = 50 gram<\/span><\/span><\/p>\n

The temperature of ice = 0\u00b0C<\/span><\/span><\/p>\n

Mass of water (mwater<\/sub>) = 200 gram<\/span><\/span><\/p>\n

Temperature of water = 30o<\/sup>C<\/span><\/span><\/p>\n

The specific heat of water (cwater<\/sub>) = 1 cal.g– 1<\/sup> \u00b0C \u20131<\/sup><\/span><\/span><\/p>\n

The heat of fusion for ice (Lice<\/sub>) = 80 cal.g \u20131<\/sup><\/span><\/span><\/p>\n

Wanted:<\/u> The final temperature<\/span><\/span><\/p>\n

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

Estimate the final condition :<\/u><\/span><\/span><\/p>\n

Heat released by water to decrease its temperature from 30o<\/sup>C to 0o<\/sup>C :<\/span><\/span><\/p>\n

Qrelease<\/sub> = mwater <\/sub>cwater<\/sub> (\u0394<\/span>T) = (200)(1)(30-0) = (200)(30) = 6000 <\/span><\/span><\/p>\n

Heat needed to melting all ice :<\/span><\/span><\/p>\n

Q = mice<\/sub> Lice<\/sub>= (50)(80) = 4000<\/span><\/span><\/p>\n

Heat used to melting all ice is 4000, while heat released by water is 6000. Can be concluded that the final temperature of the mixture above 0o<\/sup>C.<\/span><\/span><\/p>\n

Black principle :<\/span><\/span><\/p>\n

Heat released by water = heat for melting all ice + heat to increase the temperature of ice.<\/span><\/span><\/p>\n

(mwater<\/sub>)(cwater<\/sub>)(\u0394<\/span>T) = (mice<\/sub>)(Lice<\/sub>) + (mice<\/sub>)(cwater<\/sub>)(\u0394<\/span>T)<\/span><\/span><\/p>\n

(200)(1)(30-T) = (50)(80) + (50)(1)(T-0)<\/span><\/span><\/p>\n

(200)(30-T) = (50)(80) + (50)(T-0)<\/span><\/span><\/p>\n

6000 \u2013 200T = 4000 + 50T \u2013 0<\/span><\/span><\/p>\n

6000 – 4000 = 50T + 200T<\/span><\/span><\/p>\n

2000 = 250T<\/span><\/span><\/p>\n

T = 2000\/250<\/span><\/span><\/p>\n

T = 8o<\/sup>C<\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

Temperature and heat \u2013 problems and solutions 1. On a thermometer X, the freezing point of water at -30o and the boiling point of water at 90o. 60OX = \u2026.. oC. Known : The freezing point of water = -30o The boiling point of water = 90o Wanted : 60oX = ….. oC Solution :<\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Temperature and heat \u2013 problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-2104","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2104","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=2104"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2104\/revisions"}],"predecessor-version":[{"id":8647,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2104\/revisions\/8647"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2104"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2104"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2104"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}