At a certain elevation, the boiling point of water is 98.5°C. How much energy is needed to heat 35.0 mL of water to the boiling point at this elevation if the water initially was at 23.4°C?
(C_P = 75.38 J/(mol · °C) , d = 1.00 g/mL)

Question

Quizplus · Accepted Answer

First, calculate the amount of energy needed to raise the temperature of the water from 23.4°C to 98.5°C using the formula:

q1 = m x Cs x ΔT

where:
- q1 = amount of energy needed
- m = mass of water = 35.0 g
- Cs = specific heat of water = 4.18 J/(g · °C)
- ΔT = change in temperature = 98.5°C - 23.4°C = 75.1°C

Substituting the values:

q1 = 35.0 g x 4.18 J/(g · °C) x 75.1°C
q1 = 10422.85 J = 10.42 kJ

Next, calculate the amount of energy needed to vaporize the water at 98.5°C using the formula:

q2 = m x Hvap

where:
- q2 = amount of energy needed
- m = mass of water = 35.0 g
- Hvap = heat of vaporization of water at 98.5°C = 40.7 kJ/mol ÷ 18 g/mol x 1000 J/kJ = 2261.1111 J/g

Substituting the values:

q2 = 35.0 g x 2261.1111 J/g
q2 = 79139.9999 J = 79.14 kJ

Adding the two amounts of energy:

q = q1 + q2
q = 10.42 kJ + 79.14 kJ
q = 89.56 kJ

Finally, convert the answer from Joules to kilojoules:

q = 89.56 kJ

Therefore, the best choice is B) 11.0 kJ (rounded to two decimal places).

At a Certain Elevation, the Boiling Point of Water Is