A 0.259 g sample of ethanol is burnt to raise the temperature of 120 g of an oily liquid, as shown in the graph - HSC - SSCE Chemistry - Question 27 - 2013 - Paper 1

The heat transferred can be calculated using the formula:

$Q = \Delta H_{c} \times n$

Where:

• \Delta H_{c} = heat of combustion of ethanol = 1367 kJ mol⁻¹
• n = moles of ethanol = 0.00562122 mol

Thus,

$Q = 1367 kJ mol^{-1} \times 0.00562122 mol = 7.68543 kJ = 7685.43 J$

From the graph, the temperature change (∆T) can be determined by subtracting the initial temperature from the final temperature. Assuming the initial temperature is 10 °C and the final temperature is 40 °C:

$\Delta T = T_{final} - T_{initial} = 40 °C - 10 °C = 30 K$

The specific heat capacity (C) can be calculated using the formula:

$C = \frac{Q}{m \times \Delta T}$

Where:

• Q = heat transferred = 7685.43 J
• m = mass of oily liquid = 120 g = 0.120 kg
• \Delta T = 30 K

Thus,

$C = \frac{7685.43 J}{0.120 kg \times 30 K} = 2134.84 J kg^{-1} K^{-1}$

Therefore, the specific heat capacity of the oily liquid is approximately 2.1348 x 10³ J kg⁻¹ K⁻¹.