Calculate the total heat (in Joules) needed to convert 1.602 x 101 grams of ice at -1.383oC to liquid water at 8.554 x 101oC.
Melting point at 1 atm 0.0oC
Cwater(s) = 2.09 J/goC
Cwater(l) = 4.18 J/goC
delta Hofus = 6.02 kJ/mol
q1 = heat needed to raise T of solid ice at -1.383 to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)
q2 = heat needed to melt ice; i.e., change phase from solid to liquid.
q2 = mass ice x heat fusion
q3 = heat needed to raise T of liquid water from zero C to 85.54 C.
q3 = mass water x specific heat H2O x (Final-Tinitial)
Total: add q1 + q2 + q3.
To calculate the total heat needed to convert ice to liquid water, we need to consider the following steps:
1. Heating the ice from -1.383°C to 0°C.
2. Melting the ice at 0°C.
3. Heating the liquid water from 0°C to 8.554 x 10^1°C.
Let's calculate each step separately:
Step 1: Heating the ice from -1.383°C to 0°C.
The specific heat capacity of ice is assumed to be the same as water, which is Cwater(s) = 2.09 J/goC.
The heat required to raise the temperature of the ice can be calculated using the formula:
q1 = m * C * ΔT
where:
q1 = heat required for temperature change
m = mass of ice (in grams)
C = specific heat capacity of ice (in J/goC)
ΔT = change in temperature
Substituting the given values:
m = 1.602 x 10^1 grams
C = 2.09 J/goC
ΔT = 0°C - (-1.383°C) = 1.383°C
Calculating q1:
q1 = 1.602 x 10^1 grams * 2.09 J/goC * 1.383°C
q1 = 0.046 J
So, the heat required for step 1 is 0.046 J.
Step 2: Melting the ice at 0°C.
The heat required to melt the ice can be calculated using the formula:
q2 = m * ΔHfus
where:
q2 = heat required for melting
m = mass of ice (in grams)
ΔHfus = enthalpy of fusion (in J/g)
The enthalpy of fusion (ΔHfus) is given as 6.02 kJ/mol. We need to convert it into J/g:
1 mol of H2O = 18.015 g
ΔHfus = 6.02 kJ/mol = 6.02 x 10^3 J/18.015 g
Substituting the given values:
m = 1.602 x 10^1 grams
ΔHfus = 6.02 x 10^3 J/18.015 g
Calculating q2:
q2 = 1.602 x 10^1 grams * (6.02 x 10^3 J/18.015 g)
q2 = 5.34 J
So, the heat required for step 2 is 5.34 J.
Step 3: Heating the liquid water from 0°C to 8.554 x 10^1°C.
The specific heat capacity of liquid water is Cwater(l) = 4.18 J/goC.
The heat required to raise the temperature of the liquid water can be calculated using the formula:
q3 = m * C * ΔT
where:
q3 = heat required for temperature change
m = mass of liquid water (in grams)
C = specific heat capacity of liquid water (in J/goC)
ΔT = change in temperature
Substituting the given values:
m = 1.602 x 10^1 grams
C = 4.18 J/goC
ΔT = 8.554 x 10^1°C - 0°C
Calculating q3:
q3 = 1.602 x 10^1 grams * 4.18 J/goC * 8.554 x 10^1°C
q3 = 5.75 x 10^3 J
So, the heat required for step 3 is 5.75 x 10^3 J.
Finally, to find the total heat needed to convert ice to liquid water, we sum up the heats required for each step:
Total heat = q1 + q2 + q3
Total heat = 0.046 J + 5.34 J + 5.75 x 10^3 J
Total heat ≈ 5.75 x 10^3 J
Therefore, the total heat needed to convert 1.602 x 10^1 grams of ice at -1.383°C to liquid water at 8.554 x 10^1°C is approximately 5.75 x 10^3 J.
To calculate the total heat needed to convert ice at -1.383oC to liquid water at 8.554 x 101oC, we need to consider two processes: heating the ice from -1.383oC to 0.0oC (melting point) and then heating the liquid water from 0.0oC to 8.554 x 101oC.
Let's break down the calculations step by step:
1. Calculating the heat needed to heat the ice from -1.383oC to 0.0oC:
- Mass of ice, m = 1.602 x 101 grams
- Specific heat capacity of ice, Cice = 2.09 J/goC
The formula to calculate heat is Q = m * C * delta T, where Q is the heat, m is the mass, C is the specific heat capacity, and delta T is the change in temperature.
Q1 = m * Cice * delta T1
= (1.602 x 101 g) * (2.09 J/goC) * (0.0 - (-1.383)oC)
2. Calculating the heat needed to melt the ice:
- The heat of fusion for water, delta Hfus = 6.02 kJ/mol
The heat of fusion is the amount of heat required to convert one mole of a substance from a solid to a liquid state. To calculate the heat needed, we need to know the number of moles of ice.
Number of moles of ice = mass of ice / molar mass of water
= (1.602 x 101 g) / (18.015 g/mol)
Q2 = Number of moles of ice * delta Hfus
Convert kJ to J since the units should be consistent.
3. Calculating the heat needed to heat the liquid water from 0.0oC to 8.554 x 101oC:
- Mass of water, m = mass of ice
- Specific heat capacity of water, Cwater(l) = 4.18 J/goC
Q3 = m * Cwater(l) * delta T3
= (1.602 x 101 g) * (4.18 J/goC) * (8.554 x 101oC - 0.0oC)
Finally, to get the total heat needed, add up Q1, Q2, and Q3:
Total heat = Q1 + Q2 + Q3