Calculate the heat released when 10.0 g of water at 25.0°C cools to ice at 0.0°C. The specific heat of water is 1.00 cal/(g × °C); the heat of fusion is 80.0 cal/g; and the heat of vaporization is 540.0 cal/g.
how many calories are released when 25.0 g of water at 100 celcius cools to 30 celcius.
To calculate the heat released, we need to consider two steps: the heat released while cooling the water from 25.0°C to 0.0°C, and the heat released during the phase change from water to ice at 0.0°C.
Step 1: Calculating the heat released while cooling the water:
The equation for heat (q) is given by:
q = m * c * ΔT
where:
q = heat released (in calories)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in cal/(g × °C))
ΔT = change in temperature (in °C)
In this case, the mass (m) is 10.0 g, the specific heat capacity (c) is 1.00 cal/(g × °C), and the change in temperature (ΔT) is 25.0 - 0.0 = 25.0°C.
q = 10.0 g * 1.00 cal/(g × °C) * 25.0°C
q = 250 cal
Step 2: Calculating the heat released during the phase change from water to ice:
The equation for heat (q) during a phase change is given by:
q = m * ΔH
where:
q = heat released (in calories)
m = mass of the substance (in grams)
ΔH = enthalpy of the phase change (in cal/g)
In this case, the mass (m) is 10.0 g and the enthalpy of fusion (ΔH) for water is 80.0 cal/g.
q = 10.0 g * 80.0 cal/g
q = 800 cal
To find the total heat released, we add the heat released during cooling (from step 1) and the heat released during the phase change (from step 2):
Total heat released = q1 + q2
Total heat released = 250 cal + 800 cal
Total heat released = 1050 cal
Therefore, the heat released when 10.0 g of water at 25.0°C cools to ice at 0.0°C is 1050 calories.
To calculate the heat released, you need to consider the energy required to cool the water from 25.0°C to 0.0°C and the energy required for the phase change from water to ice.
First, calculate the heat required to cool the water using the specific heat formula:
Q = m * c * ΔT
Where:
Q is the heat energy
m is the mass of the water
c is the specific heat of water
ΔT is the change in temperature
Given:
m = 10.0 g
c = 1.00 cal/(g × °C)
ΔT = (0.0°C) - (25.0°C) = -25.0°C
So, plugging these values into the formula:
Q = 10.0 g * 1.00 cal/(g × °C) * (-25.0°C)
Q = -250.0 cal
Therefore, the heat required to cool the water from 25.0°C to 0.0°C is -250.0 cal.
Next, calculate the energy required for the phase change from water to ice. This is done using the formula:
Q = m * ΔHfus
Where:
Q is the heat energy
m is the mass of the water
ΔHfus is the heat of fusion
Given:
m = 10.0 g
ΔHfus = 80.0 cal/g
Plugging these values into the formula:
Q = 10.0 g * 80.0 cal/g
Q = 800.0 cal
Therefore, the heat required for the phase change from water to ice is 800.0 cal.
Finally, to calculate the total heat released, add the heat required for cooling and the heat required for the phase change:
Total heat released = -250.0 cal + 800.0 cal
Total heat released = 550.0 cal
Therefore, the heat released when 10.0 g of water at 25.0°C cools to ice at 0.0°C is 550.0 cal.
You don't need the heat of vaporization to do that calculation.
It takes 10g*1Cal/C g*10C = 100 Cal heat release to reduce the temperature ot 0C, and another 80 Cal/g*10 g = 800 Cal to freeze it. The sum is 900 Cal