what mass of ice at -14'c will be needed to cool 200cm3 of an orange drink (essentially water) from 25'c to 10'c!
(specific latent heat of fusion of ice =336000)
(specific heat capacity of ice = 2100)
(specific heat capacity of water = 4200)
I need an answer please
To calculate the mass of ice needed to cool the orange drink, we can follow these steps:
Step 1: Calculate the heat transferred from the orange drink to the ice:
Q = mcΔT
Where:
Q = Heat transferred
m = Mass
c = Specific heat capacity
ΔT = Change in temperature
First, let's calculate the heat transferred from the orange drink to cool it from 25°C to 0°C:
Q1 = (200 cm³) × (4200 J/kg°C) × (25°C - 0°C)
Q1 = 200 cm³ × 4200 J/kg°C × 25°C
Q1 = 2,100,000 J
Note: We converted the volume from cm³ to kg assuming the density of water is 1 g/cm³.
Step 2: Calculate the heat transferred from the ice to cool it from 0°C to -14°C:
Q2 = mcΔT
Where ΔT = (-14°C - 0°C) = -14°C and c = Specific heat capacity of ice (2100 J/kg°C).
We can rearrange the equation to calculate the mass (m) of ice:
m = Q2 / (c × ΔT)
m = 2,100,000 J / (2100 J/kg°C × -14°C)
m = 100 kg
Therefore, you will need 100 kg of ice at -14°C to cool the orange drink from 25°C to 0°C.
To find the mass of ice needed to cool the orange drink, we can use the equation:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's calculate the amount of heat required to cool the orange drink from 25°C to 10°C:
Q1 = (200 cm³) * (4200 J/kg°C) * (10°C - 25°C)
Q1 = (200 cm³) * (4200 J/kg°C) * (-15°C)
Q1 = -1,260,000 J
Next, let's calculate the amount of heat that would be transferred when the ice reaches its melting point:
Q2 = (m) * (336,000 J/kg)
Now, since the total amount of heat transferred is equal to the sum of Q1 and Q2, we can set up the equation:
Q1 + Q2 = 0
-1,260,000 J + (m) * (336,000 J/kg) = 0
Rearranging the equation:
(m) * (336,000 J/kg) = 1,260,000 J
Now, we can solve for the mass of ice (m):
m = (1,260,000 J) / (336,000 J/kg)
m ≈ 3.75 kg
Therefore, approximately 3.75 kg of ice at -14°C would be needed to cool 200 cm³ of the orange drink from 25°C to 10°C.