Homework Answers
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/9ed5c3f0-1a82-11ec-b353-b15e42982460.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/dcfac590-1a85-11ec-8511-239249bf122a.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/83213060-1a86-11ec-817b-7948258b439f.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/3e9ba120-1a84-11ec-949b-053a1a8b8889.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/2cbed040-1a87-11ec-b85f-03645b34d232.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](https://d2vlcm61l7u1fs.cloudfront.net/media%2F40d%2F40df40f6-b010-4c93-b9b1-a4ce6750c983%2FphpeXHtKC.png)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/cd9deea0-1a84-11ec-952d-a5e3f580861c.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/64c30870-1a87-11ec-839f-812b29c1a4ed.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/83893020-1a87-11ec-9ff5-69f5d9ac32c8.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/d41cec50-1a87-11ec-8d2a-c1c232ebf0ab.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/e7a2e220-1a87-11ec-b8ab-93f3feb48dac.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/d87cc460-1a82-11ec-b91d-b942ecea9d67.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
Answer #1
![S61 .3 Sぅx1-35..] C SH20) Сам o C ice:](http://img.homeworklib.com/questions/b8f1c980-1a86-11ec-a5ee-2b73f596121b.png?x-oss-process=image/resize,w_560)
Given Volume of water = 353 ml density of water = 1.0 g/ml mass of water = volume times density 353 times 1 = 35.3 gm we know that Specific heat of water = 4.178 j/g k heat of fusion = 335.55 j/g rightarrow Change in temperature in water = (24 - 4) = 20 degree Heat lost by water = m times s_h_20 Delta T q_lost = 353 times 4.178 times 20 = 29496.68 j rightarrow Heat gain by ice Delta T = 4 degree q_gain = m times Delta Hf + m times S_h_20 times Delta T = m times 335.55 + m times 4.178 times 4 \r\n rightarrow According to thermodynamics first law q_gain = q_lost m times 335.55 + m times 4.178 times 4 = 29496.68 (353.042) = 29496.68 m = 29496.68/352.042 m = 83.79 gm rightarrow Mass of ice melt = 83.79gm almostequalto 84 gm
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〈 Homework-Chap 11 and 12-Liquids, SOLIDS, and Intermolecular Forces Exercise 11.70 Enhanced -with Feedback 16 of...
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