How much energy (in joules) is required to convert 102 grams of ice at −38.0 °c to liquid water at 22.0 °c?

Q = 51,548.76 J

Explanation:

In this case, we need to know how much energy the ice takes to be converted in water in 22 °C.

Let's remember something, the ice passes through several stages when it's mixed with water.

In the first stage, the temperature change, but not the physical stage of ice, in other words, it remains solid.

In the second stage, the ice is reaching it's melting point and it ends when the ice is completely melted.

The third and final stage begins with the change of temperature of water once the ice is completely melted.

For these three stages, the ice is exerting a certain amount of heat. So, in order to know the energy required to convert it to water, then, we need to calculate the heat in these three stages and then do the sum of them.

Let's calculate the heat of every stage:

Q1 = mi * Ci * ΔTi

The value of Ci which is the specific heat of ice, is 2.09 J/g °C so:

Q1 = 102 * 2.09 * (0 + 38) = 8,100.84 J (a)

Now, let's calculate the heat in stage 2 which involves the enthalpy of fusion of ice (Lf) which is 334 J/g so:

Q2 = mi * Lf

Q2 = 102 * 334 = 34,068 J (b)

Finally in the thirs stage, the ice is melted, so we need to use the specific heat of water which is 4.18 J/g °C

Q3 = mi * Cw * ΔT

Q3 = 102 * 4.18 * (22 - 0) = 9,379.92 J (c)

Now that we have the heat exerted in the 3 stages, let's do the sum of the three of them:

Q = 8,100.84 + 34,068 + 9,379.92

Q = 51,548.76 J