Question

〈 Homework-Chap 11 and 12-Liquids, SOLIDS, and Intermolecular Forces Exercise 11.70 Enhanced -with Feedback 16 of 41 Part A How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 24 °C to 4 C? (Assume the density of water is 1.0 g/mL.) Express your answer using two significant figures. m= Submit Request Answer

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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