a 52g piece of ice at 0 degree is added to a sample of water at 6 degree all the ice melts and the temp of the water decreases to 0 degrees how many grams of water were in the sample?
To solve this question, we need to use the principle of energy conservation. The heat gained by the ice while melting is equal to the heat lost by the water during its temperature decrease. We can calculate the heat using the formula:
Q = m × c × ΔT
Where:
Q = Heat gained/lost
m = Mass
c = Specific heat capacity
ΔT = Temperature change
First, we need to calculate the heat gained by the ice while melting. Since all the ice melts, its temperature remains constant at 0 degrees Celsius throughout the phase change. The formula becomes:
Q_ice = m_ice × L_f
Where:
Q_ice = Heat gained by the ice
m_ice = Mass of the ice
L_f = Latent heat of fusion for ice
The latent heat of fusion for ice is 334 J/g. Therefore:
Q_ice = 52g × 334 J/g
Q_ice = 17368 J
Next, we need to calculate the heat lost by the water while decreasing from 6 degrees Celsius to 0 degrees Celsius. The specific heat capacity of water is 4.18 J/g°C. So the formula becomes:
Q_water = m_water × c_water × ΔT
Where:
Q_water = Heat lost by the water
m_water = Mass of the water (which we need to find)
c_water = Specific heat capacity of water
ΔT = Temperature change of water
We know that the water decreased from 6 degrees Celsius to 0 degrees Celsius. Therefore:
Q_water = m_water × 4.18 J/g°C × (-6°C)
Q_water = -25.08 m_water J
Now, since the heat gained is equal to the heat lost, we can set up an equation:
Q_ice = Q_water
17368 J = -25.08 m_water J
To isolate m_water, we divide both sides by -25.08 J:
m_water = 17368 J / (-25.08 J/g)
m_water ≈ -692.49 g
However, since mass cannot be negative, we know there must be an error in our calculations. Let's go back and review the calculations.
We made an assumption that the water completely freezes into ice. However, this is not possible since water reaches its maximum density at around 4 degrees Celsius. Therefore, it cannot freeze at 0 degrees without additional circumstances such as pressure changes.
Thus, without further information, it is not possible to determine the exact mass of the water sample.