A student completed an experiment to determine the amount of energy absorbed by a volume of water - HSC - SSCE Chemistry - Question 17 - 2010 - Paper 1

Using the formula for calculating heat energy:

$Q = mc\Delta T$

where:

• $Q$ is the energy absorbed (21.2 kJ = 21200 J),
• $m$ is the mass of water (0.1205 kg),
• $c$ is the specific heat capacity of water (approximately $4184 \ \text{J/kg°C}$), and
• $\Delta T$ is the change in temperature.

Rearranging the formula to solve for $\Delta T$:

$\Delta T = \frac{Q}{mc}$

Substituting the values:

$\Delta T = \frac{21200}{0.1205 \times 4184}$

Calculating $\Delta T$:

$\Delta T \approx \frac{21200}{504.658} \approx 42.0 °C$

Since the final temperature is 71.0 °C, we can find the initial temperature $T_i$:

$T_i = T_f - \Delta T = 71.0 °C - 42.0 °C = 29.0 °C$