A 135g sample of ice is in a freezer at -21 degrees celsius. It is removed from the freezer and allowed to melt and the resulting water warms up to the room temperature of 23 degrees celsius. How much energy is absorbed by the 135g sample? The specific heat capacity of ice is 2.0J gC and the heat of fusion of ice is 0.33Kj g
m=135g
T1= -21C
T2= 23 C
c= 2.0J gC
H fusion = 0.33Kj g
Q=mcT
I don't know what to do after this
q1 = heat absorbed to move ice from -21 to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinital)
q2 = heat needed to melt the ice; i.e., change solid ice @ zero C to liquid water @ zero C.
q2 = mass ice x heat fusion ice
q3 = heat needed to change T of water at zero C to 23 C.
q3 = mass water x specific heat water x (Tfinal-Tinitial)
Total Q = q1 + q2 + q3
To find the total energy absorbed by the ice during the entire process, we need to consider two steps:
Step 1: Heating the ice from -21°C to 0°C
For this step, we can use the formula:
Q1 = mcΔT
where
m = mass of the ice = 135g
c = specific heat capacity of ice = 2.0 J/g°C
ΔT = change in temperature = 0°C - (-21°C) = 21°C
Q1 = 135g * 2.0 J/g°C * 21°C
Q1 = 5670 J
Step 2: Melting the ice at 0°C to water at 0°C
To calculate the energy required for this step, we need to use the formula:
Q2 = mL
where
m = mass of the ice = 135g
L = heat of fusion of ice = 0.33 kJ/g = 0.33 * 1000 J/g
Q2 = 135g * 0.33kJ/g
Q2 = 44550 J
Step 3: Heating the water from 0°C to 23°C
Finally, we need to calculate the energy required to heat the water from 0°C to 23°C using the formula:
Q3 = mcΔT
where
m = mass of water = 135g
c = specific heat capacity of water = 4.184 J/g°C (approximately)
ΔT = change in temperature = 23°C - 0°C = 23°C
Q3 = 135g * 4.184 J/g°C * 23°C
Q3 = 135g * 96.232 J
Q3 = 12990.12 J
Total energy absorbed:
Q(total) = Q1 + Q2 + Q3
Q(total) = 5670 J + 44550 J + 12990.12 J
Q(total) = 63210.12 J
Therefore, the 135g sample of ice absorbs approximately 63210.12 J of energy during the entire process.
To calculate the total amount of energy absorbed by the 135g sample of ice, we need to consider two phases: the phase from ice to water at 0 degrees Celsius (melting), and the phase from water at 0 degrees Celsius to the final temperature at 23 degrees Celsius (warming up).
First, let's calculate the energy required to melt the ice. The formula to calculate the energy change during the phase change (melting) is:
Q_melting = m * H_fusion
Where:
Q_melting is the energy absorbed during the phase change from ice to water (in Joules)
m is the mass of the sample (in grams)
H_fusion is the heat of fusion of ice (in Joules per gram)
Plugging in the values:
m = 135g
H_fusion = 0.33KJ/g = 330J/g
Q_melting = 135g * 330J/g
Q_melting = 44,550J
So, the ice absorbs 44,550J of energy during the melting phase.
After the ice has melted, it warms up from 0 degrees Celsius to the final temperature of 23 degrees Celsius. To calculate the energy absorbed during this phase, we use the formula:
Q_warming = m * c * ΔT
Where:
Q_warming is the energy absorbed during warming (in Joules)
m is the mass of the sample (in grams)
c is the specific heat capacity of water (in Joules per gram Celsius)
ΔT is the change in temperature (in Celsius)
Plugging in the values:
m = 135g
c = 2.0J/g°C
ΔT = 23°C - 0°C = 23°C
Q_warming = 135g * 2.0J/g°C * 23°C
Q_warming = 6210J
So, the water formed from the melted ice absorbs 6210J of energy during the warming phase.
To find the total energy absorbed by the 135g sample, we need to sum the energy absorbed during melting and warming:
Total energy absorbed = Q_melting + Q_warming
Total energy absorbed = 44,550J + 6210J
Total energy absorbed = 50,760J
So, the 135g sample of ice absorbs a total of 50,760J of energy as it goes from -21°C in the freezer to 23°C at room temperature.