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

Using the heat of combustion and the moles of ethanol:

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

Where: $\Delta H_{c} = 1367 \text{ kJ mol}^{-1}$

Calculating:

$Q = 1367 \text{ kJ mol}^{-1} \times 0.005621212 \text{ mol} = 7.68543 \text{ kJ} = 7685.43 \text{ J}$

From the graph, the initial temperature (T_i) is 10.0 °C and the final temperature (T_f) is 40.0 °C. Thus, the change in temperature ( \Delta T ) is:

$\Delta T = T_{f} - T_{i} = 40.0 \text{ °C} - 10.0 \text{ °C} = 30.0 \text{ °C}$ Or in Kelvin: ( \Delta T = 30 \text{ K} )

Finally, calculate the specific heat capacity (C) using:

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

Where: m = mass of oily liquid = 120 g = 0.120 kg

Calculating:

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

Thus, the specific heat capacity of the oily liquid is approximately:

$C \approx 2.13484 \times 10^{3} \text{ J kg}^{-1} \text{ K}^{-1}$