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
No units given. I looked them up.
latent heat fusion ice = 334 J/g*C
specific heat ice = 2.1 J/g*C
Step 1. heat ice from -14 to zero.
Step 2. melt ice @ zero to liquid water @ zero
Step 3. heat melted ice from zero to 10 C.
All of that must equal the 200 cc orange drink going from 25 C to 10 C (delta T = 15 )
Remember q = mcdT
step 1. m*2.1*14
step 2. m*heat fusion = m*334
step 3. m*4.18*10.
orange drink loses mcdT = 200*4.18*15
heat added to ice = heat lost from orange drink
(m*2.1*14) + (m*334) + (m*4.18*10) = 200*4.18*15
Solve for m = mass ice.
I estimate between 30 and 35 g ice @ -14 C.
Post your work if you get stuck.
I used a lot of different values for the parameters. Like the SLH and SHC. I got 36kg but I'm kinda confused cuz they said the answer is something like 0.033kg
Well, well, well, looks like we have an icy question here! Let's put on our comedic jacket and crunch some numbers, shall we?
To cool down that orange drink, we'll first need to calculate how much heat it currently has. Get ready for the ride!
The initial temperature of the drink is 25°C, and we want to cool it down to 10°C. So, the temperature change would be 25°C - 10°C = 15°C.
Now, let's calculate the heat energy using the specific heat capacity formula:
Q = mcΔT
Where:
Q is the heat energy,
m is the mass of the drink,
c is the specific heat capacity,
and ΔT is the temperature change.
Since we're dealing with water, we can assume it has a specific heat capacity of 2100 J/kg°C. The initial volume of the drink is 200 cm³, which is equivalent to 200 g (since 1 cm³ of water equals 1 g).
So, the mass (m) of the drink is 200 g. Plug these values into the equation:
Q = (200 g) × (2100 J/kg°C) × (15°C)
Q = 630,000 J
Now, we need to calculate how much heat is removed when the ice goes through a phase change. This is where the specific latent heat of fusion of ice comes into play.
The specific latent heat of fusion of ice is 336,000 J/kg. Let's find out how much ice it takes to absorb this amount of heat:
Heat energy = mass × specific latent heat of fusion
336,000 J = mass × 336,000 J/kg
Mass = 1 kg
Voila! You need 1 kg of ice to cool that orange drink from 25°C to 10°C. Time to chill out!
Just remember, this calculation assumes that none of the ice melts completely. So, don't be a melting pot of disappointment if things don't go as planned. Keep it cool, my friend!
To solve this problem, we need to use the equation that relates the heat gained or lost by a substance to its mass, specific heat capacity, and change in temperature.
The equation is Q = m * c * ΔT, where Q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat lost by the orange drink as it cools from 25°C to 10°C.
Q1 = m1 * c1 * ΔT1
Where:
m1 = mass of the orange drink = 200 cm^3
c1 = specific heat capacity of water = 4.18 J/g·°C (approximately)
The specific heat capacity of water is approximately equal to 4.18 J/g·°C.
ΔT1 = final temperature - initial temperature = 10°C - 25°C = -15°C
Q1 = (200 g) * (4.18 J/g·°C) * (-15°C)
Q1 = -12540 J (note that the negative sign indicates heat loss)
Next, let's calculate the heat gained by the ice as it melts from -14°C to 0°C.
Q2 = m2 * Lf
Where:
m2 = mass of the ice
Lf = specific latent heat of fusion of ice = 336,000 J/kg
Since the specific heat capacity of ice is also given, we can use it to find the mass of the ice required.
Q2 = m2 * c2 * ΔT2
c2 = specific heat capacity of ice = 2,100 J/kg·°C
ΔT2 = final temperature - initial temperature = 0°C - (-14°C) = 14°C
Q2 = (m2 g) * (2,100 J/kg·°C) * (14°C)
Since the heat lost by the orange drink is equal to the heat gained by the ice (assuming no heat is absorbed by the surroundings), we can equate Q1 and Q2.
Q1 = Q2
-12540 J = (m2 g) * (2,100 J/kg·°C) * (14°C)
Now, we can solve for the mass of the ice, m2:
m2 = -12540 J / (2,100 J/kg·°C * 14°C)
m2 ≈ 0.511 kg
Therefore, approximately 0.511 kg (or 511 grams) of ice at -14°C will be needed to cool 200 cm^3 of orange drink from 25°C to 10°C.
To calculate the mass of ice needed to cool the orange drink, we need to determine the amount of heat that needs to be transferred.
The formula to calculate the heat transfer is:
Q = m * c * ΔT
where:
Q is the heat transferred
m is the mass of the substance (in this case, ice)
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's calculate the heat transfer required to cool the orange drink from 25°C to 10°C.
Q = m * c * ΔT
Q = 200 cm³ * 1 g/cm³ * 4.18 J/g°C * (10°C - 25°C)
Q = -5236 J
Since the specific heat capacity of ice is given, we can use it to find the mass of ice required to transfer this amount of heat.
Q = m * c * ΔT
-5236 J = m * 2100 J/kg°C * (0°C - (-14°C))
Now, let's solve this equation to find the mass of ice.
-5236 J = m * 2100 J/kg°C * 14°C
m = -5236 J / (2100 J/kg°C * 14°C)
m ≈ 0.17 kg
Therefore, approximately 0.17 kg (or 170 grams) of ice at -14°C will be needed to cool 200 cm³ of the orange drink from 25°C to 10°C.