What quantity of heat is necessary to convert 50.0 g of ice at 0.0 C into steam 100,0 C? The heat of fusion is 80.0 cal/g, the heat of vaporization is 540 cal/g, and the specific heat of water is 1.00 cal/gC.
q1 = heat needed to melt ice at zero C to liquid water at zero C.
q1 = mass ice x heat fusion
q2 = heat needed to raise T from zero C to 100 C.
q2 = mass melted ice x specific heat water x (Tfinal-Tinitial)
q3 = heat needed to vaporize water at 100 to steam at 100.
q3 = mass water x heat vaporization.
Total Q = q1 + q2 + q3.
To determine the quantity of heat required to convert the ice into steam, we need to consider three steps: heating the ice from 0.0 °C to 0.0 °C, melting the ice into water at 0.0 °C, and finally heating the water from 0.0 °C to 100.0 °C and converting it into steam.
Step 1: Heating the ice from 0.0 °C to 0.0 °C
The specific heat of ice is 1.00 cal/g°C. Since the temperature doesn't change, no heat is gained or lost during this step.
Step 2: Melting the ice into water at 0.0 °C
The heat of fusion of ice is 80.0 cal/g. Therefore, the heat required to melt the ice can be calculated as follows:
Heat = mass × heat of fusion
Heat = 50.0 g × 80.0 cal/g = 4000 cal
Step 3: Heating the water from 0.0 °C to 100.0 °C and converting it into steam
The specific heat of water is 1.00 cal/g°C. Therefore, the heat required to heat the water can be calculated as follows:
Heat = mass × specific heat × change in temperature
Heat = 50.0 g × 1.00 cal/g°C × (100.0 °C - 0.0 °C) = 5000 cal
The total heat required is the sum of the heat from each step:
Total heat = Heat in Step 1 + Heat in Step 2 + Heat in Step 3
Total heat = 0 cal + 4000 cal + 5000 cal = 9000 cal
Therefore, 9000 calories of heat is necessary to convert 50.0 g of ice at 0.0 °C into steam at 100.0 °C.
To calculate the quantity of heat required to convert the given amount of ice at 0.0°C into steam at 100.0°C, we need to consider the different stages of the phase change and the specific heat capacities involved.
First, we need to calculate the heat required to heat the ice from 0.0°C to its melting point at 0.0°C. The specific heat capacity of ice is 1.00 cal/g°C. The formula for calculating the heat required is:
Q1 = mass * specific heat capacity * temperature change
Q1 = 50.0 g * 1.00 cal/g°C * (0.0°C - 0.0°C)
Q1 = 0.0 cal
Since there is no temperature change, no heat is required to raise the ice's temperature to its melting point. The melting of ice requires energy, called the heat of fusion, to break the intermolecular bonds. The formula for calculating the heat required for the phase change from ice to water is:
Q2 = mass * heat of fusion
Q2 = 50.0 g * 80.0 cal/g
Q2 = 4000 cal
Next, we need to calculate the heat required to heat the resulting water from 0.0°C to its boiling point at 100.0°C. The specific heat capacity of water is also 1.00 cal/g°C. The formula for calculating the heat required is:
Q3 = mass * specific heat capacity * temperature change
Q3 = 50.0 g * 1.00 cal/g°C * (100.0°C - 0.0°C)
Q3 = 5000 cal
Lastly, we need to calculate the heat required for the phase change from water to steam. The heat of vaporization is 540 cal/g. The formula for calculating the heat required is:
Q4 = mass * heat of vaporization
Q4 = 50.0 g * 540 cal/g
Q4 = 27000 cal
To determine the total quantity of heat required to convert the ice at 0.0°C into steam at 100.0°C, we sum up the individual quantities of heat:
Total heat = Q1 + Q2 + Q3 + Q4
Total heat = 0.0 cal + 4000 cal + 5000 cal + 27000 cal
Total heat = 36000 cal
Therefore, a total of 36000 calories of heat is required to convert 50.0 grams of ice at 0.0°C into steam at 100.0°C.