What mass of ice is needed to cool 60g of water from 43oC to 20oC
yujh
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To calculate the mass of ice needed to cool water, we can use the equation:
Heat gained by ice = Heat lost by water
The heat lost by water can be calculated using the equation:
Q = mcΔT
Where:
Q is the heat lost by water (in joules)
m is the mass of water (in grams)
c is the specific heat capacity of water (4.18 J/(g⋅°C))
ΔT is the change in temperature of water (final temperature - initial temperature)
The water temperature changes from 43°C to 20°C, therefore ΔT = 20°C - 43°C = -23°C.
Now let's calculate the heat lost by water:
Q = mcΔT
Q = (60g) * (4.18 J/(g⋅°C)) * (-23°C)
Q = -5799.6 J
The heat gained by the ice can be calculated using the formula:
Q = mL
Where:
Q is the heat gained by the ice (in joules)
m is the mass of ice (in grams)
L is the latent heat of fusion of water (334 J/g)
Since the ice is at 0°C, it needs to be warmed up to 0°C first:
Q1 = mcΔT
Q1 = (mice) * (4.18 J/(g⋅°C)) * (0°C - (-23°C))
Q1 = 966.14 m
Next, the ice needs to melt:
Q2 = mL
Q2 = mice * 334 J/g
The total heat gained by the ice is the sum of Q1 and Q2:
Qgained = Q1 + Q2
Qgained = 966.14 m + (mice * 334 J/g)
Since the heat gained by the ice must be equal to the heat lost by the water, we can set these two equations equal to each other:
Qgained = Qlost
966.14 m + (mice * 334 J/g) = -5799.6 J
Now we can solve for the mass of the ice (mice):
966.14 m + 334 mice = -5799.6
334 mice = -6765.74 - 966.14 m
mice = (-6765.74 - 966.14 m) / 334
Given the mass of water (m = 60 g), we can substitute this into the equation to obtain the mass of ice (mice):
mice = (-6765.74 - 966.14 * 60) / 334
mice = -6765.74 - 57968.4 / 334
mice = -6765.74 - 173.84
mice ≈ 6939.58 g
Therefore, approximately 6939.58 grams (or 6.93958 kg) of ice is needed to cool 60 grams of water from 43°C to 20°C.
To find the mass of ice needed to cool the water, we need to use the formula:
Q = m * c * ΔT
Where:
Q = heat transfer (in this case, it represents the heat that needs to be removed from the water to cool it down)
m = mass of the substance (in this case, the mass of ice we want to find)
c = specific heat capacity (which is 4.18 J/g°C for water)
ΔT = change in temperature (final temperature - initial temperature)
First, let's calculate the heat transfer for the water using its specific heat capacity:
Q_water = m_water * c_water * ΔT_water
m_water = 60 g (given)
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_water = 43°C - 20°C = 23°C
Q_water = 60 g * 4.18 J/g°C * 23°C
Q_water = 58596 J
Now, we know that the heat transfer (Q) from the water should be equal to the heat transfer absorbed by the ice:
Q_ice = m_ice * c_ice * ΔT_ice
Since the ice starts at 0°C and ends at 20°C:
ΔT_ice = 20°C - 0°C = 20°C
We also know that the specific heat capacity of ice (c_ice) is 2.09 J/g°C.
Now we can substitute the known values and solve for m_ice:
Q_ice = m_ice * c_ice * ΔT_ice
58596 J = m_ice * 2.09 J/g°C * 20°C
Dividing both sides of the equation by (2.09 J/g°C * 20°C) gives us:
m_ice = 58596 J / (2.09 J/g°C * 20°C)
m_ice = 1401.67 g
Therefore, approximately 1401.67 grams of ice are needed to cool 60 grams of water from 43°C to 20°C.