Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
Global warming refers to the rise in average global temperature due to the increased concentration of certain gases, called gre...Asked by marie
Global warming refers to the rise in average global temperature due to the increased concentration of certain
gases, called greenhouse gases, in our atmosphere. Earth¡¯s oceans, because of their high heat capacity, can absorb
heat and therefore act to slow down global warming. How much heat would be required to warm Earth¡¯s
oceans by 1.0 ¡ãC? Assume that the volume of Earth¡¯s oceans is 137 x ¡¼10¡½^7 ¡¼km¡½^3 and that the density of seawater
is 1.03 g/cm3. Also assume that the heat capacity of seawater is the same as that of water.
gases, called greenhouse gases, in our atmosphere. Earth¡¯s oceans, because of their high heat capacity, can absorb
heat and therefore act to slow down global warming. How much heat would be required to warm Earth¡¯s
oceans by 1.0 ¡ãC? Assume that the volume of Earth¡¯s oceans is 137 x ¡¼10¡½^7 ¡¼km¡½^3 and that the density of seawater
is 1.03 g/cm3. Also assume that the heat capacity of seawater is the same as that of water.
Answers
Answered by
caleb
137 x 10^7 km^3
1.03 g / cm^3
4.184 g / cm^3
We need to convert the volume of the oceans to cm^3 so density (g/cm^3 is in the same units.
1 km = 1000 meters
1 meter = 39.37 inches
1 inch = 2.54 cm
1 km x (1000 meters / 1km) = 1000 meters
1000 meters x (39.37inches / 1 meter) = 39,370 inches
39,370 inches x (2.54 cm / 1 inch) = 99999.8 centimeters
so, 1 km = 99999.8 cm but we need the conversion factor for cubic km and cubic cm so we cubed these conversion factors now.
(1km) ^3 = (99999.8 cm) ^3 = 1^3 km^3 = 99999.8 ^3 cm^3 = 1 km^3 = 9.99994 x 10^14 cm^3
We now use this last conversion in our original data.
(137 x 10^7 km^3) x (9.99994 x 10^14 cm^3 / 1 km) = 1.37 x 10^24 cubic centimeters
(1.37 x 10^24 cubic centimeters) x (1.03g / cubic centimeters) = 1.41 x 10^24 grams of water in the world's oceans
(1.41 x 10^24 grams) (4.184 joules / g degrees celsius) grams cancel and you are left with joules per degree celsius
= 5.90 x 10^24 joules / degree celsius. The real answer is how many joules are required to raise the temperature of one gram of water one degree Celsius, so this
answers our question. This question came from a Nivaldo Tro book that I have and the answer is not listed at the back of the book. Hope this helps.
1.03 g / cm^3
4.184 g / cm^3
We need to convert the volume of the oceans to cm^3 so density (g/cm^3 is in the same units.
1 km = 1000 meters
1 meter = 39.37 inches
1 inch = 2.54 cm
1 km x (1000 meters / 1km) = 1000 meters
1000 meters x (39.37inches / 1 meter) = 39,370 inches
39,370 inches x (2.54 cm / 1 inch) = 99999.8 centimeters
so, 1 km = 99999.8 cm but we need the conversion factor for cubic km and cubic cm so we cubed these conversion factors now.
(1km) ^3 = (99999.8 cm) ^3 = 1^3 km^3 = 99999.8 ^3 cm^3 = 1 km^3 = 9.99994 x 10^14 cm^3
We now use this last conversion in our original data.
(137 x 10^7 km^3) x (9.99994 x 10^14 cm^3 / 1 km) = 1.37 x 10^24 cubic centimeters
(1.37 x 10^24 cubic centimeters) x (1.03g / cubic centimeters) = 1.41 x 10^24 grams of water in the world's oceans
(1.41 x 10^24 grams) (4.184 joules / g degrees celsius) grams cancel and you are left with joules per degree celsius
= 5.90 x 10^24 joules / degree celsius. The real answer is how many joules are required to raise the temperature of one gram of water one degree Celsius, so this
answers our question. This question came from a Nivaldo Tro book that I have and the answer is not listed at the back of the book. Hope this helps.
Answered by
caleb
I apologize but in the original data at the top of my answer the heat capacity should read:
(4.184 joules / (gram x degree celsius))
(4.184 joules / (gram x degree celsius))
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.