Answer:
The amount of thermal energy needed is 15167500 joules.
Explanation:
By First Law of Thermodynamics, we see that amount of thermal energy ([tex]Q[/tex]), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components ([tex]U_{l,ice}[/tex], [tex]U_{l,steam}[/tex]), in joules, and two sensible component ([tex]U_{s,w}[/tex]), in joules, that is:
[tex]Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam}[/tex] (1)
By definitions of Sensible and Latent Heat, we expanded the formula:
[tex]Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w})[/tex] (2)
Where:
[tex]m[/tex] - Mass, in kilograms.
[tex]h_{f,w}[/tex] - Latent heat of fussion of water, in joules per kilogram.
[tex]h_{v,w}[/tex] - Latent heat of vaporization of water, in joules per kilogram.
[tex]c_{ice}[/tex] - Specific heat of ice, in joules per kilogram per degree Celsius.
[tex]c_{w}[/tex] - Specific heat of water, in joules per kilogram per degree Celsius.
[tex]\Delta T_{ice}[/tex] - Change in temperature of ice, measured in degrees Celsius.
[tex]\Delta T_{w}[/tex] - Change in temperature of water, measured in degrees Celsius.
If we know that [tex]m = 5\,kg[/tex], [tex]h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}[/tex], [tex]h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}[/tex], [tex]c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}[/tex], [tex]c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}[/tex], [tex]\Delta T_{ice} = 10\,^{\circ}C[/tex] and [tex]\Delta T_{w} = 100\,^{\circ}C[/tex], then the amount of thermal energy is:
[tex]Q = 15167500\,J[/tex]
The amount of thermal energy needed is 15167500 joules.
