The avg moler heat capacities of ice and water are 37.8 and 75.6jmol-1 respectively and the enthalpy of fusion of ice is 6.012kjmol-1 the amount of heat required
to change 10 gm of ice at -10°c to water at10°c would be
Well, let's do some calculations and have a little fun with it!
First, we need to find out the amount of heat required to change 10 grams of ice at -10°C to water at 10°C. Now, we know that the enthalpy of fusion of ice is 6.012 kJ/mol, but we need to find the heat required for 10 grams of ice.
Since the molar mass of water is roughly 18 g/mol (H2O has two hydrogens with a molar mass of 1 g/mol each and one oxygen with a molar mass of 16 g/mol), we can divide 10 grams by the molar mass to get the number of moles:
10 g / 18 g/mol = approximately 0.556 moles.
Now, we can multiply the number of moles by the enthalpy of fusion of ice:
0.556 moles * 6.012 kJ/mol = approximately 3.34 kJ.
So, the amount of heat required to change 10 grams of ice at -10°C to water at 10°C would be approximately 3.34 kJ. Just be careful not to melt any frozen jokes in the process!
To calculate the amount of heat required to change 10 grams of ice at -10°C to water at 10°C, we need to consider the following steps:
1. Calculate the heat required to raise the temperature of the ice from -10°C to 0°C:
q1 = mass * specific heat capacity * change in temperature
= 10 g * 37.8 J mol^(-1) K^(-1) * (0 - (-10))°C
= 3780 J
2. Calculate the heat required to change the ice at 0°C to water at 0°C:
q2 = enthalpy of fusion * number of moles
= 6.012 kJ mol^(-1) * (10 g / molar mass of water)
= 6.012 kJ * (10 g / 18.015 g mol^(-1))
= 3.337 kJ
3. Calculate the heat required to raise the temperature of the water from 0°C to 10°C:
q3 = mass * specific heat capacity * change in temperature
= 10 g * 75.6 J mol^(-1) K^(-1) * (10 - 0)°C
= 7560 J
4. Sum up the individual heats to get the total heat required:
heat required = q1 + q2 + q3
= 3780 J + 3.337 kJ + 7560 J
= 11537 J
Therefore, the amount of heat required to change 10 grams of ice at -10°C to water at 10°C is approximately 11537 Joules.
To calculate the amount of heat required to change 10 grams of ice at -10°C to water at 10°C, we need to consider the following steps:
1. Calculate the heat required to warm up the ice from -10°C to its melting point (0°C).
2. Calculate the heat required to melt the ice at 0°C.
3. Calculate the heat required to warm up the resulting water from 0°C to 10°C.
Step 1: Calculate the heat required to warm up the ice
The specific heat capacity of ice (c_ice) is given as 37.8 J/mol°C. First, we need to convert the mass of ice from grams to moles using the molar mass of water (H2O), which is approximately 18 g/mol.
10g of ice = (10g) / (18g/mol) ≈ 0.556 mol
Now, calculate the heat required using the equation:
q_1 = (mass) * (specific heat capacity) * (change in temperature)
q_1 = (0.556 mol) * (37.8 J/mol°C) * (0°C - (-10°C))
q_1 = (0.556 mol) * (37.8 J/mol°C) * (10°C)
q_1 ≈ 211.848 J
Step 2: Calculate the heat required to melt the ice
The enthalpy of fusion (∆H_fusion) of ice is given as 6.012 kJ/mol. Convert it to joules per mole:
∆H_fusion = 6.012 kJ/mol = 6.012 * 10^3 J/mol
Now, calculate the heat required using the equation:
q_2 = (moles) * (enthalpy of fusion)
q_2 = (0.556 mol) * (6.012 * 10^3 J/mol)
q_2 ≈ 3.34 kJ = 3.34 * 10^3 J
Step 3: Calculate the heat required to warm up the resulting water
The specific heat capacity of water (c_water) is given as 75.6 J/mol°C. We need to convert the mass of water to moles using the molar mass of water (H2O), which is approximately 18 g/mol.
10g of water = (10g) / (18g/mol) ≈ 0.556 mol
Now, calculate the heat required using the equation:
q_3 = (moles) * (specific heat capacity) * (change in temperature)
q_3 = (0.556 mol) * (75.6 J/mol°C) * (10°C - 0°C)
q_3 = (0.556 mol) * (75.6 J/mol°C) * (10°C)
q_3 ≈ 416.456 J
Now, we can calculate the total heat required by adding up the heat required in each step:
Total heat required = q_1 + q_2 + q_3
Total heat required ≈ 211.848 J + 3.34 * 10^3 J + 416.456 J
Total heat required ≈ 3.967 * 10^3 J
q1 = heat to change T of ice @ -10 C to ice @ zero.
q1 = mass ice x specific heat ice x (T1-T2) = ? T2 is 0 and T1 is -10
q2 = heat to change solid water (ice) to liquid water (liquid)
q2 = mass ice x heat fusion = ?
q3 = heat needed to change liquid water @ 0 to liquid at +10.
q3 = mass water x specific heat water x (T2 - T1) = ?
qtotal = q1 + q2 + q3 = ?