How many grams of ice are needed to cool 250 ml of water from 75 degrees celcius to 5 degrees celcius?
Note the correct spelling of celsius.
Heat gained by ice + heat gained by melted water + heat lost by original water = 0
[mass ice x heat fusion] + [mass melted ice x specific water x (Tfinal-Tinitial)] + [mass 250mL water x specific heat water x (Tfinal-Tinitial)] = 0
Solve for mass ice (and melted ice). I think something like 200 g ice is necessary but that is an aproximation.
To find the amount of ice needed to cool water, you 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
ΔT is the change in temperature
Assuming the density of water is approximately 1 g/ml, we can calculate the mass of water using the equation:
Mass = Volume * Density
Mass = 250 ml * 1 g/ml
Mass = 250 g
Since ice has a specific heat capacity of 2.09 J/g°C and we're starting at 75°C and want to cool to 5°C, the change in temperature is:
ΔT = 75°C - 5°C
ΔT = 70°C
Now we can calculate the amount of heat transferred:
Q = mcΔT
Q = 250 g * 4.184 J/g°C * 70°C
Q = 730,100 J
Since the heat transferred Q is equal to the heat absorbed by the ice, which is equal to the mass of the ice multiplied by its specific heat capacity, we can rearrange the equation to solve for the mass of the ice:
Q = mc
m = Q / c
m = 730,100 J / 2.09 J/g°C
m ≈ 349,760 g
Therefore, approximately 349,760 grams (or 349.76 kg) of ice are needed to cool 250 ml of water from 75 degrees Celsius to 5 degrees Celsius.
To determine the number of grams of ice needed to cool 250 ml of water from 75 degrees Celsius to 5 degrees Celsius, we need to consider the concept of heat transfer.
The heat transfer equation, Q = m * c * ΔT, can be used to calculate the heat gained or lost during a temperature change. Here:
Q is the heat transferred (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C), and
ΔT is the change in temperature (in °C).
In this scenario, we want to calculate the mass of ice (m) required to cool water from 75°C to 5°C. First, let's find the heat lost by the water using the heat transfer equation:
Q(water) = m(water) * c(water) * ΔT(water)
The specific heat capacity of water (c) is approximately 4.18 J/g°C. Let's assume the initial temperature of the water is 75°C, and we want to cool it to 5°C. In this case:
Q(water) = m(water) * 4.18 J/g°C * (75 - 5)°C
To determine the mass of water (m(water)) in grams, we need to consider the density of water. The density of water is approximately 1 g/ml. Therefore:
m(water) = 250 ml * 1 g/ml
Now, we can substitute the known values into our equation to find the heat lost by the water (Q(water)):
Q(water) = (250 ml * 1 g/ml) * 4.18 J/g°C * (75 - 5)°C
Now that we know the heat lost by the water, we can determine the heat gained by the ice. When ice melts, it absorbs heat without a change in temperature. The heat transferred can be calculated using the equation:
Q(ice) = m(ice) * ΔH(fusion)
ΔH(fusion) is the enthalpy of fusion, which represents the heat required to convert a substance from a solid to a liquid at its melting point. For water, ΔH(fusion) is approximately 333.5 J/g. We assume the ice is initially at its melting point, which is 0°C.
Q(ice) = m(ice) * 333.5 J/g
Since the heat lost by the water is equal to the heat gained by the ice, we can set the equations equal to each other:
Q(water) = Q(ice)
(m(water) * 4.18 J/g°C * (75 - 5)°C) = (m(ice) * 333.5 J/g)
Now we can solve for m(ice), the mass of ice needed:
m(ice) = ((m(water) * 4.18 J/g°C * (75 - 5)°C) / 333.5 J/g)
By substituting the known values, we can find the mass of ice (m(ice)) required to cool 250 ml of water from 75°C to 5°C.