What mass of ice at minus 14 degree Celsius will be needed to cool 200cm raise to power 3 of an orange drink (essentially) what mass of water from 25 degree Celsius to 10 centigrade? (Specific latent heat of fusion of ice is 3.36 raise to power minus ten square five joules per kg. Specific heat capacity of water 4200 joules kilogram raise to power minus one per kelvin raise to power minus one. Specific heat capacity of ice= 2100kg-¹K-1
To calculate the mass of ice needed to cool the orange drink or the mass of water that needs to be cooled, we will first use the formula:
Q = mcΔT
where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. For the orange drink, we will use the specific heat capacity of water, and for ice, we will use the specific heat capacity of ice.
Let's start by calculating the mass of ice required to cool the 200 cm³ of orange drink.
Step 1: Calculate the change in temperature for the orange drink.
The initial temperature of the orange drink is not given, so we will assume it to be the same as its final temperature, which is -14°C. The change in temperature (ΔT) would then be:
ΔT = final temperature - initial temperature
ΔT = -14°C - initial temperature
Step 2: Convert the change in temperature to kelvin.
To convert the change in temperature in Celsius to kelvin, add 273 to the temperature in Celsius:
ΔT (in Kelvin) = ΔT (in Celsius) + 273
Step 3: Substitute the values into the formula.
Using the specific heat capacity of water (c = 4200 J/(kg·K)), we can rearrange the formula to solve for mass:
m = Q / (c ΔT)
Given:
V = 200 cm³ (which is equivalent to 200 g, assuming the density of the orange drink is the same as water)
Using the above information, we can substitute the values into the formula:
m = (200 g) × (4200 J/(kg·K)) × (ΔT (in Kelvin))
Step 4: Calculate the mass by converting from grams to kilograms.
Convert the mass from grams to kilograms by dividing by 1000:
m (in kg) = m (in g) / 1000
Now, let's calculate the mass of ice needed to cool the orange drink.
First, convert the change in temperature:
ΔT = -14°C - initial temperature
Assuming the initial temperature of the orange drink is 25°C, the change in temperature is:
ΔT = -14 - 25 = -39°C
Convert the change in temperature to Kelvin by adding 273:
ΔT (in Kelvin) = -39 + 273 = 234 K
Substitute the values into the formula:
m = (200 g) × (4200 J/(kg·K)) × (234 K)
m = 19,752,000 J
Since the specific latent heat of fusion of ice (L) is given as 3.36x10^5 J/kg, we can use the formula:
Q = mL
To calculate the mass of ice needed, rearrange the formula:
m = Q / L
Given:
Q = 19,752,000 J
L = 3.36x10^5 J/kg
Substitute the values:
m = 19,752,000 J / (3.36x10^5 J/kg)
m = 58.8 kg
Therefore, you will need approximately 58.8 kg of ice at -14°C to cool 200 cm³ (or 200 g) of the orange drink.
Note: The specific heat capacity of ice (c = 2100 J/(kg·K)) is not required for this specific calculation as we are using the specific latent heat of fusion of ice (L) to determine the ice mass needed.