Define specific latent heat - Leaving Cert Physics - Question c - 2014

To calculate the mass of the ice, we can use the formula:

ho V $$ Where: - $m$ is the mass of the ice, - $\rho$ is the density of ice (0.92 g/cm³), - $V$ is the volume of the ice cubes. The volume of one cube of ice is: $$ V = s^3 = (2.5 ext{ cm})^3 = 15.625 ext{ cm}^3 $$ Since there are three cubes: $$ V_{total} = 3 imes 15.625 ext{ cm}^3 = 46.875 ext{ cm}^3 $$ Now converting to m³ (1 cm³ = 1 x 10⁻⁶ m³): $$ V_{total} = 46.875 imes 10^{-6} ext{ m}^3 $$ Calculating mass: $$ m = 0.92 imes (46.875 imes 10^{-6}) = 43.125 ext{ g} $$

We will use the energy balance principle:

The heat gained by the ice is equal to the heat lost by the water:

$Q_{ice} = Q_{water}$

Breaking it down:

1. Heating the ice from -20 °C to 0 °C: $Q_{ice ext{, when warming}} = m_{ice} c_{ice} riangle T$
   Where:

   • $m_{ice} = 0.043125 ext{ kg}$,
   • $c_{ice} = 2100 ext{ J kg}^{-1} ext{ K}^{-1}$,
   • $\Delta T = 0 - (-20) = 20 ext{ K}$.

   Thus, $Q_{ice ext{, warming}} = 0.043125 imes 2100 imes 20$ $= 1811.25 ext{ J}$

2. Melting the ice: $Q_{ice ext{, melting}} = m_{ice} L_{f}$
   Where:

   • $L_{f} = 3.3 imes 10^5 ext{ J kg}^{-1}$.

   Thus, $Q_{ice ext{, melting}} = 0.043125 imes 3.3 imes 10^5$ $= 14231.25 ext{ J}$

3. Heating the water: $Q_{water} = m_{water} c_{water} (T_{final} - 24)$ where:

   • $m_{water} = 0.5 ext{ kg}, c_{water} = 4200 ext{ J kg}^{-1} ext{ K}^{-1}$.

Combining: $1811.25 + 14231.25 = 0.5 imes 4200 imes (T - 24)$

Solving for $T$: $18112.5 = 2100(T - 24)$ $T - 24 = \frac{18112.5}{2100}$ $T = 24 + 8.61428571 ext{ °C}$ $T = 16.06 ext{ °C}$