How many grams of ice are needed to cool 250 ml of water from 75 degrees Celsius to 5 degrees Celsius?
(m_ice * 6.01kJ) + (m_ice * 4.18J/g°C * (5-75°C) = -[250mL * 4.18J/g°C * (5-75°C)] ???
Most is ok but that second term I think should be mice*4.18*(5-0). At any rate, I would change the 6.01 kJ/mol to J/g*C = 334 J/g*C
Then [mice*334] + [mice*4.18*(5-0)] + [250 x 4.18 x (5-75) = 0
I solved and got close to 206 g ice needed.
Ah, trying to calculate the amount of ice needed to cool water, I see. Well, I'd say your equation looks a bit icy, let's give it a little warmth, shall we?
First, let's correct the units. Since your ice melting and water cooling are happening at the same temperature, we don't need to use the heat of fusion for the ice (6.01 kJ/g). Instead, we can focus on the specific heat capacity of ice, which is approximately 2.09 J/g°C.
Now, let's rearrange the equation a bit, using the correct units:
(m_ice * 2.09 J/g°C * (75-5)°C) = -[250 mL * 4.18 J/g°C * (5-75)°C]
Simplifying further:
m_ice * 2.09 J/g°C * 70°C = -250 mL * 4.18 J/g°C * (-70)°C
Now we divide by the values:
m_ice = [-250 mL * 4.18 J/g°C * (-70)°C] / (2.09 J/g°C * 70°C)
And if we do the math, we get:
m_ice ≈ 125 g of ice
So, approximately 125 grams of ice should be enough to cool your water from 75 degrees Celsius to 5 degrees Celsius. Just remember, ice to meet you at the chilly rendezvous!
To determine the number of grams of ice needed to cool 250 ml of water from 75 degrees Celsius to 5 degrees Celsius, we can use the principle of heat transfer between the ice and the water.
First, we need to find the heat absorbed by the ice. This can be calculated using the heat formula:
Q = m * c * ΔT
Where:
Q is the heat absorbed or released (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C),
ΔT is the change in temperature (in °C).
Given that the water is being cooled from 75°C to 5°C, we get:
ΔT = 5 - 75 = -70°C
The specific heat capacity of ice (c) is approximately 2.09 J/g°C.
So, if we let m_ice represent the mass of ice needed, we have:
(m_ice * 2.09 J/g°C * -70°C) = (250 ml * 4.18 J/g°C * -70°C)
Since 1 gram of water equals 1 ml, we can convert the volume of water (250 ml) to grams by assuming a density of 1 g/ml for water:
(m_ice * 2.09 J/g°C * -70°C) = (250 g * 4.18 J/g°C * -70°C)
Simplifying the equation further:
(m_ice * -146.3 J) = (-67,150 J)
Dividing both sides by -146.3 J:
m_ice = (-67,150 J) / (-146.3 J)
m_ice ≈ 459.29 grams
Therefore, approximately 459.29 grams of ice are needed to cool 250 ml of water from 75 degrees Celsius to 5 degrees Celsius.
To calculate the amount of ice needed to cool the water, you need to use the equation:
(Mass of ice * Heat of fusion) + (Mass of ice * Specific heat capacity of ice * change in temperature) = -(Mass of water * Specific heat capacity of water * change in temperature)
Let's break down the equation and solve it step by step.
1. Find the change in temperature for the water:
The initial temperature of the water is 75°C, and the final temperature is 5°C. So, the change in temperature is:
Change in temperature = Final temperature - Initial temperature
Change in temperature = 5°C - 75°C = -70°C
2. Calculate the specific heat capacity of water:
The specific heat capacity of water is approximately 4.18 J/g°C.
3. Convert the volume of water to its mass:
The volume of water is given as 250 ml. Since the density of water is 1 g/ml, the mass of water is:
Mass of water = Volume of water * Density of water
Mass of water = 250 ml * 1 g/ml = 250 g
4. Substitute the values into the equation:
( Mass of ice * Heat of fusion ) + ( Mass of ice * Specific heat capacity of ice * Change in temperature ) = -( Mass of water * Specific heat capacity of water * Change in temperature )
Let's assume the heat of fusion of ice is 6.01 kJ/kg. To convert it into J/g, divide by 1000:
Heat of fusion = 6.01 kJ/kg = 6.01 kJ/g = 6010 J/g
Let's consider the term on the right-hand side of the equation as well:
Mass of water * Specific heat capacity of water * Change in temperature = -( Mass of water * Specific heat capacity of water * -70°C )
5. Simplify the equation and solve for the mass of ice:
(Mass of ice * 6010 J/g) + (Mass of ice * 2.09 J/g°C * 70°C) = - (250 g * 4.18 J/g°C * -70°C)
Combine like terms:
6010M + 146.3M = - ( 250 * 4.18 * 70 )
Simplify further:
6216.3M = 625,150
Divide both sides by 6216.3:
M = 625,150 / 6216.3
M ≈ 100.60 g
Therefore, approximately 100.60 grams of ice are needed to cool 250 ml of water from 75°C to 5°C.