Question

Using the heat of fusion for water 334 J/g the heat of vaporization for water 2260 J / g and fhe specific heat of water 4.184 J/g C calculate the total amount of heat for each of the following

Using the heat of fusion for water, 334 J/g. the heat of vaporication for water 2260 J/6, and the specific heat of water, 4 184J/g °C, calculate the total amount of heat for each of the following Part A joules released when 1.75g of water cools from 17.0 to 0 °C and freezes Express your answer to three significant figures. IVO AXO ? J Submit Request Answer Part B joules needed to melt 400 g of ice af 'Cand to warm the liquid to 700 "C Express your answer to three significant figures. VALO O ? J Submit Request Answer Part klojoules released when 110 g of steam condenses af 100 °C and the liquid cools to 090 Express your answer to three significant figures. IVO AE o ? kJ Submit Regwest Answer

Answer 1

A)

Ti = 17.0 oC

Tf = 0.0 oC

Cl = 4.184 J/g.oC

Heat released to convert liquid from 17.0 oC to 0.0 oC

Q1 = m*Cl*(Ti-Tf)

= 1.75 g * 4.184 J/g.oC *(17-0) oC

= 124.474 J

Hf = 334 J/g

Heat released to convert liquid to solid at 0.0 oC

Q2 = m*Hf

= 1.75g *334 J/g

= 584.5 J

Total heat released = Q1 + Q2

= 124.474 J + 584.5 J

= 708.974 J

Answer: 709 J

B)

Ti = 0.0 oC

Tf = 70.0 oC

here

Hf = 334 J/g

Heat required to convert solid to liquid at 0.0 oC

Q1 = m*Hf

= 40.0g *334 J/g

= 13360 J

Cl = 4.184 J/g.oC

Heat required to convert liquid from 0.0 oC to 70.0 oC

Q2 = m*Cl*(Tf-Ti)

= 40 g * 4.184 J/g.oC *(70-0) oC

= 11715.2 J

Total heat required = Q1 + Q2

= 13360 J + 11715.2 J

= 25075.2 J

Answer: 25100 J

C)

Ti = 100.0 oC

Tf = 0.0 oC

Hv = 2260 J/g

Heat released to convert gas to liquid at 100.0 oC

Q1 = m*Hv

= 11.0g *2260 J/g

= 24860 J

Cl = 4.184 J/g.oC

Heat released to convert liquid from 100.0 oC to 0.0 oC

Q2 = m*Cl*(Ti-Tf)

= 11 g * 4.184 J/g.oC *(100-0) oC

= 4602.4 J

Total heat released = Q1 + Q2

= 24860 J + 4602.4 J

= 29462 J

= 29.5 KJ

Answer: 29.5 KJ

D)

Ti = 0.0 oC

Tf = 100.0 oC

here

Hf = 334 J/g

Heat required to convert solid to liquid at 0.0 oC

Q1 = m*Hf

= 23.5g *334 J/g

= 7849 J

Cl = 4.184 J/g.oC

Heat required to convert liquid from 0.0 oC to 100.0 oC

Q2 = m*Cl*(Tf-Ti)

= 23.5 g * 4.184 J/g.oC *(100-0) oC

= 9832.4 J

Hv = 2260 J/g

Heat required to convert liquid to gas at 100.0 oC

Q3 = m*Hv

= 23.5g *2260 J/g

= 53110 J

Total heat required = Q1 + Q2 + Q3

= 7849 J + 9832.4 J + 53110 J

= 70791.4 J

= 70.8 KJ

Answer: 70.8 KJ