Two Equations Needed

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**Problem #1:** How many kJ are required to heat 45.0 g of H_{2}O at 25.0 °C and then boil it all away?

**Solution:**

Comment: We must do two calculations and then sum the answers.

1) The first calculation uses this equation:

q = (mass) (Δt) (C_{p})This summarizes the information needed:

Δt = 75.0 °C

The mass = 45.0 g

C_{p}= 4.184 J g¯^{1}°C¯^{1}

2) Substituting, we have:

q = (45.0 g) (75.0 °C) (4.184 J g¯^{1}°C¯^{1})q = 14121 J = 14.121 kJ

3) The second calculation uses this equation:

q = (moles of water) (ΔH_{vap})This summarizes the information needed:

ΔH_{vap}= 40.7 kJ/mol

The mass = 45.0 g

The molar mass of H_{2}O = 18.0 gram/mol

4) Substituting, we obtain:

q = (45.0 g / 18.0 g mol¯^{1}) (40.7 kJ/mol)q = 101.75 kJ

5) Adding:

101.75 kJ + 14.121 kJ = 116 kJ (to three sig figs)

**Problem #2:** How many kJ need to be removed from a 120.0 g sample of water, initially at 25.0 °C, in order to freeze it at 0 °C? (Area three, then area two on the time-temperature graph.)

**Solution:**

1) The first calculation:

q = (mass) (Δt) (C_{p})q = (120.0 g) (25.0 °C) (4.184 J g¯

^{1}°C¯^{1})q = 12,552 J = 12.552 kJ

2) The second calculation:

q = (moles of water) (ΔH_{vap})q = (120.0 g / 18.0 g mol¯

^{1}) (6.02 kJ/mol)q = 40.13 kJ

3) Summing up the values from the two steps gives 52.8 kJ.

**Problem #3:** You are given 12.0 g of ice at -5.00 °C. How much energy is needed to melt the ice completely to water?

**Solution:**

1) The first calculation:

q = (mass) (Δt) (C_{p})q = (12.0 g) (5.0 °C) (2.06 J g¯

^{1}°C¯^{1})q = 123.6 J = 0.1236 kJ

2) The second calculation:

q = (moles of water) (ΔH_{vap})q = (12.0 g / 18.0 g mol¯

^{1}) (6.02 kJ/mol)q = 4.0133 kJ

3) Summing up the values from the two steps gives 4.14 kJ, to three significant figures.

**Problem #4:** Lead has a melting point of 327.5 °C, its specific heat is 0.128 J/g **⋅**°C, and its molar enthalpy of fusion is 4.80 kJ/mol. How much heat, in kilojoules, will be required to heat a 500.0 g sample of lead from 23.0 °C to its melting point and then melt it?

**Solution:**

1) Two calculations are required:

a) heat iron from 23.0 to 327.5

b) melt iron at 327.5

2) Here are the calculation set-ups:

q_{a}= (500.0 g) (304.5 °C) (0.128 J/g⋅°C) = 19488 J

q_{b}= (500.0 g / 55.845 g/mol) (4.80 kJ/mol) = 42.976 kJ

3) Add:

19.488 kJ + 42.976 kJ = 62.5 kJ (to three sig figs)Note that I changed 19488 J to kJ before adding.

**Problem #5:** The specific heat capacity of silver is 0.235 J/g-K. Its melting point is 962.0 °C, and its enthalpy of fusion is 11.3 kJ/mol. What quantity of energy, in Joules, is required to change 9.10 g of silver from a solid at 25.0 °C to a liquid at 962 °C?

**Solution:**

1) Two calculations are required:

a) heat silver from 25.0 to 962

b) melt silver at 962

2) Here are the calculation set-ups:

q_{1}= (9.10 g) (937.0 K) (0.235 J/g-K) = 2003.77 J

q_{2}= (9.10 g / 107.87 g/mol) (11.3 kJ/mol) = 0.953277 kJ = 953.277 J

3) The answer:

2003.77 J + 953.277 J = 2957.047 JNote how I use 937.0 K. This is because it is a difference, not an actual temperature. The difference between 962.0 °C and 25.0 °C is 937.0 K. You can see this by converting the two Celsius values to their kelvin values and then subtracting. You'll get 937.0 K for the difference.To three sig figs, 2960 J

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