113k views
5 votes
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)

1 Answer

2 votes

Answer:

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.

User Manny Fitzgerald
by
7.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.