How much heat must be removed from 15.0g of water at 10°C to become ice at 0°C?
150 cal to take the water to 0º C
79.7 cal/g to freeze it
To find out how much heat must be removed from the water, we need to calculate the heat energy required to change the temperature of the water from 10°C to 0°C and then convert it to ice.
The heat energy required to change the temperature of a substance can be calculated using the formula:
q = m * c * ΔT
Where:
q is the heat energy (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
For water, the specific heat capacity is 4.18 J/g°C, assuming it is in liquid form.
First, let's calculate the heat energy required to cool the water from 10°C to 0°C:
q1 = m * c * ΔT1
q1 = 15.0g * 4.18 J/g°C * (0°C - 10°C)
q1 = 15.0g * 4.18 J/g°C * (-10°C)
q1 = -627 J
Note: The negative sign indicates that heat is being removed from the system (water) as it is cooling down.
Next, let's calculate the heat energy required to convert the water at 0°C to ice at 0°C. This process is known as the heat of fusion.
The heat of fusion for water is 333.55 J/g.
q2 = m * ΔHf
q2 = 15.0g * 333.55 J/g
q2 = 5003.25 J
Finally, to find the total heat energy required, we add q1 and q2:
Total heat energy required = q1 + q2
= -627 J + 5003.25 J
= 4376.25 J
Therefore, approximately 4376.25 J of heat must be removed from 15.0g of water at 10°C to become ice at 0°C.