An ice tray with 175 g of water at 30°C is placed in a freezer at –9°C. How much energy must be removed from
the water to lower its temperature to the freezing point?
heat out = specific heat of water * mass of water * (30-0)
0.25kcal/kg-C
To calculate the amount of energy that must be removed from the water to lower its temperature to the freezing point, you can use the specific heat capacity formula.
The specific heat capacity of water is 4.18 J/g°C.
Step 1: Calculate the temperature change
The initial temperature of the water is 30°C, and the freezing point is 0°C. The temperature change is 30°C - 0°C = 30°C.
Step 2: Calculate the energy required
Energy = mass * specific heat capacity * temperature change
The mass of water is given as 175 g.
The specific heat capacity of water is 4.18 J/g°C.
The temperature change is 30°C.
Energy = 175 g * 4.18 J/g°C * 30°C
Energy = 21915 J (rounded to the nearest whole number)
Therefore, approximately 21915 J (rounded to the nearest whole number) of energy must be removed from the water to lower its temperature to the freezing point.
To calculate the amount of energy that must be removed from the water to lower its temperature to the freezing point, you can use the equation:
Q = mcΔT
Where:
Q is the amount of energy transferred (in Joules)
m is the mass of the water (in grams)
c is the specific heat capacity of water (4.184 J/g°C)
ΔT is the change in temperature (in °C)
Step 1: Convert the mass of water from grams to kilograms.
175 g = 0.175 kg
Step 2: Calculate the change in temperature.
ΔT = Tfinal - Tinitial
Tfinal = 0 °C (freezing point of water)
Tinitial = 30 °C
ΔT = 0 °C - 30 °C = -30 °C
Note that we include the negative sign to indicate a decrease in temperature.
Step 3: Substitute the values into the equation to calculate the amount of energy removed.
Q = (0.175 kg)(4.184 J/g°C)(-30 °C)
Q = -22.092 J
The negative sign indicates that energy is being removed from the water. The absolute value of the energy is 22.092 J.
Therefore, the amount of energy that must be removed from the water to lower its temperature to the freezing point is approximately 22.092 Joules.