Question

What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to

Answer 1

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy ($Q$), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components ($U_{l,ice}$, $U_{l,steam}$), in joules, and two sensible component ($U_{s,w}$), in joules, that is:

$Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam}$ (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

$Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w})$ (2)

Where:

$m$ - Mass, in kilograms.

$h_{f,w}$ - Latent heat of fussion of water, in joules per kilogram.

$h_{v,w}$ - Latent heat of vaporization of water, in joules per kilogram.

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

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

$\Delta T_{ice}$ - Change in temperature of ice, measured in degrees Celsius.

$\Delta T_{w}$ - Change in temperature of water, measured in degrees Celsius.

If we know that $m = 5\,kg$, $h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}$, $h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}$, $c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$, $c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$, $\Delta T_{ice} = 10\,^{\circ}C$ and $\Delta T_{w} = 100\,^{\circ}C$, then the amount of thermal energy is:

$Q = 15167500\,J$

The amount of thermal energy needed is 15167500 joules.