What is the minimum amount of ice (at 0 degress C) that must be added to 250mL of water at 23 degrees C in order to cool it to 0 degrees celsius? (Density of water= .98 g/cm^3; heat of fusion= 333J/g; heat capacity of water= 4.184J/g C; and heat capacity of ice= 1.86 J/g C)
heat gained by ice + heat lost by water = 0
(mass ice x heat fusion) + [mass water x specific heat water x (Tfinal-Tinitial)]
To determine the minimum amount of ice required to cool the water to 0 degrees Celsius, we need to calculate the heat energy lost by the water and the heat energy gained by the ice as they reach thermal equilibrium.
Step 1: Calculate the heat energy lost by the water:
The heat energy lost by the water can be calculated using the equation:
Q = m * c * ΔT
where Q is the heat energy lost, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
m = 250 mL = 250 g (since the density of water is 0.98 g/cm^3)
c = 4.184 J/g °C
ΔT = 23 °C (since the water needs to cool to 0 °C)
Plugging in the values:
Q = 250 g * 4.184 J/g °C * 23 °C
Q = 24406 J
Step 2: Calculate the heat energy gained by the ice:
The heat energy gained by the ice can be calculated using the equation:
Q = m * c * ΔT
where Q is the heat energy gained, m is the mass of ice, c is the specific heat capacity of ice, and ΔT is the change in temperature.
Given:
c (specific heat capacity of ice) = 1.86 J/g °C
ΔT = 0 °C (since the ice is at 0 °C)
We need to solve for m, the mass of ice.
Plugging in the values:
24406 J = m * 1.86 J/g °C * 0 °C
This equation shows that no heat energy is gained by the ice, as there is no change in temperature (ΔT = 0 °C). Therefore, the mass of ice required is 0 grams.
Hence, no ice is needed to cool the water from 23 °C to 0 °C since the water loses heat energy and the ice gains no heat energy.
To determine the minimum amount of ice that must be added to cool the water to 0 degrees Celsius, we need to calculate the heat exchange that occurs during the cooling process.
The heat exchange can be calculated using the formula:
Q = m * c * ΔT
Where Q is the heat exchange, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat exchange for the water. The mass of water can be determined using its density:
Density of water = mass / volume
Rearranging the equation, we get:
Mass of water = density of water * volume of water
Mass of water = 0.98 g/cm^3 * 250 mL
Note that we need to convert mL to cm^3 since 1 mL is equivalent to 1 cm^3.
Mass of water = 0.98 g/cm^3 * 250 cm^3
Now, let's calculate the heat exchange for the water using its specific heat capacity:
Q_water = mass of water * specific heat capacity of water * ΔT
ΔT = final temperature - initial temperature
ΔT = 0°C - 23°C
ΔT = -23°C
Substituting the values into the equation:
Q_water = (0.98 g/cm^3 * 250 cm^3) * (4.184 J/g°C) * (-23°C)
Now, let's calculate the heat exchange for the ice using its specific heat capacity:
Q_ice = mass of ice * specific heat capacity of ice * ΔT
Since the ice is initially at 0 degrees Celsius, its final temperature is also 0 degrees Celsius.
Q_ice = mass of ice * specific heat capacity of ice * ΔT
Q_ice = mass of ice * (1.86 J/g°C) * 0°C
To cool the water to 0 degrees Celsius, the heat gained by the ice must be equal to the heat lost by the water:
Q_water = Q_ice
(0.98 g/cm^3 * 250 cm^3) * (4.184 J/g°C) * (-23°C) = mass of ice * (1.86 J/g°C) * 0°C
Simplifying the equation:
(0.98 g/cm^3 * 250 cm^3) * (4.184 J/g°C) * (-23°C) = mass of ice * 0 J
Now, solve the equation for the mass of ice:
mass of ice = [(0.98 g/cm^3 * 250 cm^3) * (4.184 J/g°C) * (-23°C)] / 0 J
mass of ice = [(0.98 g/cm^3 * 250 cm^3) * (4.184 J/g°C) * (-23°C)] / 0 g
Simplifying further:
mass of ice = 0 g (since anything divided by zero is undefined)
Therefore, the minimum amount of ice that must be added to cool the water to 0 degrees Celsius is zero grams.