Question

Question 16

8 Points

Silver is known to be an excellent conductor of heat and electricity. if 8.20 kJ of energy are provided, calculate the energy required (Joules - just put in the number, 4 sig figs) to heat 14.25 g of silver from 25.00 °C to its melting point BLANK-1, the energy required (kJ - just put in the number, 4 sig figs) to melt the silver BLANK-2, and determine whether the silver will reach the boiling point. (yes/no) BLANK-3. What will the final temperature of the silver be (^oC - just put in the number, 4 sig figs) BLANK-4? (Do not include commas in your answers, give all numeric answers to 4 significant figures, do not use scientific notation)

{heat capacity of solid silver = 24.80 J K^-1 mol^-1, heat capacity of liquid silver 33.40 J K^-1 mol^-1, heat capacity of gaseous silver 21.10 J K^-1 mol^-1, ΔH_fus (Ag) = 11.97 kJ/mol, ΔH_vap (Ag) = 250.6 kJ/mol, m.p. Ag = 962.0° C, b.p. Ag = 2212° C }

• BLANK-1
• BLANK-2
• BLANK-3
• BLANK-4

Answer 1

1b)

Part 1

Calculation for heat required to increase the temperature of silver from 25⁰ to its melting point (962⁰)

heat capacity of Silver = 24.80 J K^-1 mol^-1

moles of Silver = ( mass / molar mass) = ( 14.25/107.8) = 0.132

H₁ = moles of silver $\times$ heat capacity of solid silver $\times$ temperature change

= 0.132 $\times$ 24.80 $\times$ ( 962 - 25)

= 3067 J

= 3.067 KJ

answer of Blank 1 = 3.067 KJ

Part 2

Calculation for heat required to melt silver

H₂ = moles of Silver $\times$ Enthalpy of fusion of Silver

= 0.132 (mol) $\times$ 11.97 (KJ/mol)

= 1.580 KJ

Answer of BLANK 2 is 1.580 KJ

Part 3

Total heat remains = 8.20 KJ - ( H₁ +H₂) = 8.25 - (3.067 + 1.580) = 3.602 KJ

now heat required to reach boiling point of liquid silver

H₃ = moles of silver $\times$ heat capacity of liquid silver $\times$ temperature change

= 0.132  $\times$ 33.40 $\times$ ( 2212 - 962)

= 5511 J

= 5.511 KJ

As heat required to raise the temperature to boiling point is less than heat remain after melting of silver ,  so it will not reach boiling point.

Therefore answer of BLANK 3 is NO.

part 4

Heat = moles of silver $\times$ heat capacity of liquid silver $\times$ temperature change

or .3.602 $\times$ 1000 (J ) = 0.132 $\times$ 33.40 $\times$ ( T_f - 962)

or 3602 = 4.4088 $\times$ ( T_f - 962)

or, ( T_f - 962) = 817

or, T_f = 1779 ⁰c

Hence final temperature is 1779 ⁰c.

answer of BLANK 4 is  1779 ⁰C.