How many kilojoules of energy are needed to convert 104 g of ice at -14.6 to water at 25.8°C? (The specific heat of ice at -14.6 is 2.01 J/g°C.)
q1 = head needed to raise T of solid ice at -14.6 (is that C?) to zero.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)
q2 = heat needed to melt ice.
q2 = mass ice x heat fusion
q3 = heat needed to raise T of water from zero C to 25.8 C.
q3 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Total Q = q1 + q2 + q3
Thank you Dr Bob
To answer this question, we need to calculate the energy required to raise the temperature of the ice from -14.6°C to 0°C and then convert it from ice at 0°C to water at 0°C, and finally raise the temperature of water from 0°C to 25.8°C.
Step 1: Calculate the energy required to heat the ice from -14.6°C to 0°C.
The specific heat capacity (C) of ice at -14.6°C is given as 2.01 J/g°C. The mass (m) of ice is 104 g. The temperature change (ΔT) is 0°C - (-14.6°C) = 14.6°C.
The energy (Q) required to heat the ice is calculated using the formula:
Q = m * C * ΔT
Substituting the given values, we get:
Q = 104 g * 2.01 J/g°C * 14.6°C
Q = 2996.64 J
Therefore, it takes 2996.64 J of energy to heat the ice from -14.6°C to 0°C.
Step 2: Calculate the energy required to melt the ice from 0°C to water at 0°C.
The heat of fusion (ΔHfus) for ice is 334 J/g. Since all the ice is expected to melt at 0°C, we can directly calculate the energy required.
The energy (Q) required to melt the ice is calculated using the formula:
Q = m * ΔHfus
Substituting the given values, we get:
Q = 104 g * 334 J/g
Q = 34736 J
Therefore, it takes 34736 J of energy to melt 104 g of ice at 0°C to water at 0°C.
Step 3: Calculate the energy required to heat the water from 0°C to 25.8°C.
The specific heat capacity (C) of water is generally considered to be 4.18 J/g°C. The mass (m) of the water is 104 g. The temperature change (ΔT) is 25.8°C - 0°C = 25.8°C.
The energy (Q) required to heat the water is calculated using the formula:
Q = m * C * ΔT
Substituting the given values, we get:
Q = 104 g * 4.18 J/g°C * 25.8°C
Q = 10919.88 J
Therefore, it takes 10919.88 J of energy to heat 104 g of water from 0°C to 25.8°C.
Step 4: Calculate the total energy required.
To find the total energy required, we add up the energies calculated in steps 1, 2, and 3:
Total energy = Energy to heat ice + Energy to melt ice + Energy to heat water
Total energy = 2996.64 J + 34736 J + 10919.88 J
Total energy = 48652.52 J
Therefore, it takes 48652.52 J (or 48.65 kJ) of energy to convert 104 g of ice at -14.6°C to water at 25.8°C.