Calculate the enthalpy change for converting 10.0g of ice at -25 degrees Celsius to water at 80 degrees Celsius. The specific heats of ice, water, and steam are 2.09j/g degrees Celsius, 4.18j/g degrees Celsius, and 1.85j/g degrees Celsius. The enthalpy of fusion for water is 6.01kj/mol and the enthalpy evaporation is 40.67kj/mol. How do I solve this?

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To solve this problem, we need to break it down into a few steps:

1. Calculate the enthalpy change required to heat the ice from -25 degrees Celsius to 0 degrees Celsius.
2. Calculate the enthalpy change required to melt the ice at 0 degrees Celsius to water at 0 degrees Celsius.
3. Calculate the enthalpy change required to heat the water from 0 degrees Celsius to 80 degrees Celsius.
4. Add up the enthalpy changes calculated in steps 1, 2, and 3 to get the total enthalpy change.

Let's go through each step in detail:

Step 1: Calculate the enthalpy change to heat the ice:

First, we need to calculate the heat required to raise the temperature of the ice from -25 degrees Celsius to 0 degrees Celsius. To do this, we will use the specific heat capacity of ice.

The formula to calculate the heat required is: q = m * C * ΔT
where q is the heat, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Given:
Mass (m) of ice = 10.0 g
Specific heat capacity (C) of ice = 2.09 J/g-°C
Change in temperature (ΔT) = 0°C - (-25°C) = 25°C

Plugging in the values, we get:
q = 10.0 g * 2.09 J/g-°C * 25°C
= 522.5 J

So, the heat required to raise the temperature of the ice from -25°C to 0°C is 522.5 J.

Step 2: Calculate the enthalpy change to melt the ice:

Now, we need to calculate the heat required to melt the ice at 0 degrees Celsius into water at 0 degrees Celsius. This can be calculated using the enthalpy of fusion.

The formula to calculate the heat required is: q = m * ΔH_fusion
where q is the heat, m is the mass, and ΔH_fusion is the enthalpy of fusion.

Given:
Mass (m) of ice = 10.0 g
Enthalpy of fusion (ΔH_fusion) = 6.01 kJ/mol

First, let's calculate the number of moles of ice:
Number of moles (n) = mass (m) / molar mass (M)
The molar mass of water is approximately 18.015 g/mol, so:
n = 10.0 g / 18.015 g/mol
= 0.554 mol

Next, we can calculate the heat required to melt the ice:
q = 0.554 mol * 6.01 kJ/mol
= 3.33 kJ
= 3330 J

So, the heat required to melt the ice is 3330 J.

Step 3: Calculate the enthalpy change to heat the water:

Now, we need to calculate the heat required to raise the temperature of the water from 0 degrees Celsius to 80 degrees Celsius. To do this, we will use the specific heat capacity of water.

The formula to calculate the heat required is: q = m
* C * ΔT
where q is the heat, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Given:
Mass (m) of water = 10.0 g
Specific heat capacity (C) of water = 4.18 J/g-°C
Change in temperature (ΔT) = 80°C - 0°C = 80°C

Plugging in the values, we get:
q = 10.0 g * 4.18 J/g-°C * 80°C
= 3344 J

So, the heat required to raise the temperature of the water from 0°C to 80°C is 3344 J.

Step 4: Add up the enthalpy changes:

Finally, we add up the enthalpy changes calculated in steps 1, 2, and 3 to get the total enthalpy change.

Total enthalpy change = heat required to heat the ice + heat required to melt the ice + heat required to heat the water
= 522.5 J + 3330 J + 3344 J
= 7196.5 J

The total enthalpy change for converting 10.0 g of ice at -25 degrees Celsius to water at 80 degrees Celsius is 7196.5 J.

To solve this problem, we need to consider the different steps involved in the conversion of ice at -25 degrees Celsius to water at 80 degrees Celsius. Each step involves a change in enthalpy, which we can calculate using the specific heat capacities and enthalpies provided.

Step 1: Heating the ice from -25 degrees Celsius to 0 degrees Celsius.
This step involves raising the temperature of the ice without changing its phase. We can calculate the enthalpy change using the formula:

Q = m * C * ΔT

Where:
Q = heat energy
m = mass
C = specific heat capacity
ΔT = change in temperature

In this case, the specific heat capacity of ice (C) is 2.09 J/g °C and the change in temperature (ΔT) is 0 - (-25) = 25 degrees Celsius.

Q1 = 10.0g * 2.09 J/g °C * 25 °C
Q1 = 522.5 J

Step 2: Melting the ice at 0 degrees Celsius to water at 0 degrees Celsius (enthalpy of fusion).
This step involves changing the phase of the substance from solid to liquid. We can calculate the enthalpy change using the formula:

Q = m * ΔH_fusion

Where:
Q = heat energy
m = mass
ΔH_fusion = enthalpy of fusion

In this case, the mass (m) is still 10.0g, and the enthalpy of fusion (ΔH_fusion) is given as 6.01 kJ/mol. We need to convert kJ to J, and since we're given the enthalpy per mole, we need to convert grams to moles using the molar mass of water (H2O).

The molar mass of water (H2O) = 18.01528 g/mol (2 * 1.008 + 16.00).

Q2 = 10.0g * (6.01 kJ/mol * 1000 J/kJ) / (18.01528 g/mol)
Q2 = 3.34 kJ * 1000 J/kJ = 3340 J

Step 3: Heating the water from 0 degrees Celsius to 80 degrees Celsius.
This step involves raising the temperature of the water without changing its phase. Similar to step 1, we can use the formula:

Q = m * C * ΔT

In this case, the specific heat capacity of water (C) is 4.18 J/g °C and the change in temperature (ΔT) is 80 - 0 = 80 degrees Celsius.

Q3 = 10.0g * 4.18 J/g °C * 80 °C
Q3 = 3344 J

The total enthalpy change is the sum of Q1, Q2, and Q3:

Enthalpy change = Q1 + Q2 + Q3
= 522.5 J + 3340 J + 3344 J
= 7206.5 J

Therefore, the enthalpy change for converting 10.0g of ice at -25 degrees Celsius to water at 80 degrees Celsius is 7206.5 J.