What mass of ice at -14 degree Celsius will be needed to cool 200 cm of an orange drink from 25 degree Celsius to 10 degree? (specific latent heat of fusion of ice=336000,specific heat capacity of ice=2100 specific heat capacity of water=4200
To calculate the mass of ice needed to cool the orange drink, you need to consider the energy transfer during the phase change when the ice melts and the subsequent temperature change.
1. Calculate the energy needed to cool the orange drink:
First, calculate the energy to cool the orange drink from 25°C to 0°C:
Q1 = mass_orange * specific_heat_orange * (final_temp_orange - initial_temp_orange)
Q1 = 200 g * 4200 J/kg°C * (10°C - 25°C)
Q1 = -420,000 J
2. Calculate the energy needed to melt the ice:
The energy required for the phase change (from solid to liquid) is given by:
Q2 = mass_ice * specific_latent_heat_fusion
Q2 = mass_ice * 336,000 J/kg
3. Calculate the energy needed to heat the water:
The energy required to heat the melted ice (water) from 0°C to 10°C is:
Q3 = mass_water * specific_heat_water * (final_temp_water - initial_temp_water)
Q3 = mass_water * 4200 J/kg°C * (10°C - 0°C)
Q3 = 42,000 J
Since the total energy required (Q total) for the cooling process is equal to the sum of Q1, Q2, and Q3, we can write:
Q total = Q1 + Q2 + Q3
However, Q2 is negative because it represents the heat released during the phase change. So, we can rewrite the equation as:
Q total = Q1 - |Q2| + Q3
4. Solve for the mass of ice:
Let's substitute the given values and solve for the mass_ice:
Q1 - |Q2| + Q3 = 0
-420,000 J - (mass_ice * 336,000 J/kg) + 42,000 J = 0
-mass_ice * 336,000 J/kg = 420,000 J - 42,000 J
mass_ice = (378,000 J) / (336,000 J/kg)
mass_ice = 1.125 kg (or 1125 grams)
Therefore, you will need 1125 grams (or 1.125 kg) of ice at -14°C to cool 200 cm³ of orange drink from 25°C to 10°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 exchanged
m is the mass
c is the specific heat capacity
Δt is the change in temperature
First, we need to calculate the amount of heat exchanged when cooling the orange drink. We can use the following equation:
Q1 = mcΔt
Where:
Q1 is the heat exchanged when cooling the orange drink
m is the mass of the orange drink
c is the specific heat capacity of water
Δt is the change in temperature of the orange drink
Q1 = (200 cm)(4200 J/kg°C)(25°C - 10°C)
= (200 g)(4200 J/kg°C)(15°C)
= 1260000 J
Next, we need to calculate the amount of heat transferred during the phase change from solid ice to liquid water. We can use the equation:
Q2 = mLf
Where:
Q2 is the heat transferred during the phase change
m is the mass of ice
Lf is the specific latent heat of fusion of ice
Q2 = mLf
= (mass of ice)(336000 J/kg)
Since the ice is initially at -14°C, it needs to be heated to 0°C and undergo phase change from solid to liquid.
Q2 = (mass of ice)(2100 J/kg°C)(0°C - (-14°C)) + (mass of ice)(336000 J/kg)
= (mass of ice)(2100 J/kg°C)(14°C) + (mass of ice)(336000 J/kg)
Now, we can equate the total heat exchanged to the sum of Q1 and Q2:
Q1 + Q2 = 1260000 J + (mass of ice)(2100 J/kg°C)(14°C) + (mass of ice)(336000 J/kg)
We can rearrange the equation to solve for the mass of ice:
mass of ice = (Q1 + (mass of ice)(2100 J/kg°C)(14°C)) / (336000 J/kg - 2100 J/kg°C(14°C))
We can now substitute the values into the equation and solve for the mass of ice:
mass of ice = (1260000 J + (mass of ice)(2100 J/kg°C)(14°C)) / (336000 J/kg - 2100 J/kg°C(14°C))
This is a nonlinear equation that requires an iterative numerical method to solve. However, it is difficult to solve manually. Therefore, computational methods or software are usually used to solve such equations.
Your nonuse of units on the heats is disappointing. I am going to guess they are in J/kg, even in that, they are rouned loosely.
The sum heat changes = zero.
-massice*2100J/kg-massice*336000J/kg -(massice)4200(10-0)+200(4200)(25-10)=0
check that. Negative means heat absorbed, positive means heat given off.