A student used the apparatus below to determine the molar heat of combustion of propanol. The following results were obtained: Mass of 1-propanol burnt = 0.60 g Mass of water heated = 200 g Initial temperature of water = 21.0°C The molar heat of combustion of 1-propanol is 2021 kJ mol⁻¹. Assuming no heat loss, what would be the final temperature of the water? (A) 24.2°C (B) 29.1°C (C) 45.2°C (D) 48.4°C

SSCE HSC Chemistry Alcohols: A student used the apparatus bel

A student used the apparatus below to determine the molar heat of combustion of propanol - HSC - SSCE Chemistry - Question 13 - 2004 - Paper 1

Question 13

A student used the apparatus below to determine the molar heat of combustion of propanol. The following results were obtained: Mass of 1-propanol burnt = 0.60 g Ma... show full transcript

Worked Solution & Example Answer:A student used the apparatus below to determine the molar heat of combustion of propanol - HSC - SSCE Chemistry - Question 13 - 2004 - Paper 1

Step 1

Determine the heat released from combustion

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Answer

First, calculate the heat released by the combustion of 1-propanol. The heat, Q, can be calculated using the formula:

$Q = n imes ext{molar heat of combustion}$

where n is the number of moles of 1-propanol burnt. The molar mass of 1-propanol (C₃H₈O) is approximately 60 g/mol. Thus,

$n = \frac{0.60 ext{ g}}{60 ext{ g/mol}} = 0.01 ext{ mol}$

Now, substituting the values into the heat formula:

$Q = 0.01 ext{ mol} \times 2021 ext{ kJ/mol} = 20.21 ext{ kJ} = 20210 ext{ J}$

Step 2

Calculate the final temperature of water

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Answer

Next, we can find the final temperature of the water using the heat absorbed by it. The heat absorbed by the water can be calculated using:

$Q = m imes c imes \Delta T$

where:

m = mass of water = 200 g
c = specific heat capacity of water = 4.18 J/g°C
\Delta T = final temperature - initial temperature

Thus,

$20210 ext{ J} = 200 ext{ g} \times 4.18 ext{ J/g°C} \times (T_f - 21.0°C)$

Now, calculate \Delta T:

$20210 = 836 \times (T_f - 21.0)$

$T_f - 21.0 = \frac{20210}{836} ≈ 24.2°C$

Therefore,

$T_f = 24.2 + 21.0 ≈ 45.2°C$

A student used the apparatus below to determine the molar heat of combustion of propanol - HSC - SSCE Chemistry - Question 13 - 2004 - Paper 1

Worked Solution & Example Answer:A student used the apparatus below to determine the molar heat of combustion of propanol - HSC - SSCE Chemistry - Question 13 - 2004 - Paper 1

Determine the heat released from combustion

Calculate the final temperature of water

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