Question

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

steam at 100 °C? (Heat of fusion of Ice is 3.34 x 10 j/kg, specific heat capacity of
ice is 2.09 x 10 3 J/kg °C, Specific heat capacity of water is 4.186 x 10 ? J/kg° C,
Heat of vaporization of water 2.26 x 10 J/kg)

Answer 1

The amount of thermal energy needed is 15167500 joules.

Step-by-step 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* 10^(5)\,(J)/(kg)`,
`h_(v,w) = 2.26* 10^(6)\,(J)/(kg)`,
`c_(ice) = 2.090* 10^(3)\,(J)/(kg\cdot ^(\circ)C)`,
`c_(w) = 4.186* 10^(3)\,(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.