Question

Calculate the total energy (in kJ) absorbed when 50.5 g of ice at -15.0°C is converted into liquid water at 65.0 °C.

Answer 1

Answer:

The total energy absorbed is 32.171 kilojoules.

Explanation:

The total energy absorbed by the ice is the sum of the sensible heat of ice and water and the latent heat of fusion of the water, that is:

$Q = m\cdot [c_{i}\cdot (T_{2}-T_{1})+L_{f} + c_{w}\cdot (T_{3}-T_{2})]$ (1)

Where:

$m$ - Mass of the ice, in kilograms.

$c_{i}$ - Specific heat of ice, in kilojoules per kilogram-degree Celsius.

$c_{w}$ - Specific heat of water, in kilojoules per kilogram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in kilojoules per degree Celsius.

$T_{1}$ - Initial temperature of water, in degrees Celsius.

$T_{2}$ - Fusion point of water, in degrees Celsius.

$T_{3}$ - Final temperature of water, in degree Celsius.

$Q$ - Total energy absorbed, in kilojoules.

If we know that $m = 50.5\times 10^{-3}\,kg$, $c_{i} = 2.090 \,\frac{kJ}{kg\cdot^{\circ}C}$, $c_{w} = 4.180\,\frac{J}{kg\cdot ^{\circ}C}$, $L_{f} = 334\,\frac{kJ}{kg}$, $T_{1} = -15\,^{\circ}C$, $T_{2} = 0\,^{\circ}C$ and $T_{3} = 65\,^{\circ}C$, then the total energy absorbed is:

$Q= (50.5\times 10^{-3}\,kg)\cdot \left[\left(2.090\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (15\,^{\circ}C) + 334\,\frac{kJ}{kg}+ \left(4.180\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (65\,^{\circ}C)\right]$$Q = 32.171\,kJ$

The total energy absorbed is 32.171 kilojoules.