A 49g piece of ice at 0.0C is added to a sample of water at 9.0C. All of the ice melts and the temperature of the water decreases to 0.0C . How many grams of water were in the sample?
To solve this problem, we need to use the concept of heat transfer and the principle of conservation of energy.
First, let's calculate the heat required to melt the ice. The heat required to melt the ice can be calculated using the formula:
Q1 = m1 * ΔHf
Where:
Q1 is the heat required to melt the ice (in Joules)
m1 is the mass of the ice (in grams)
ΔHf is the heat of fusion for water, which is 334 J/g
Substituting the given values:
m1 = 49g
ΔHf = 334 J/g
Q1 = 49g * 334 J/g
Q1 = 16366 J
Next, let's calculate the heat lost by the water. The heat lost by the water can be calculated using the formula:
Q2 = m2 * Cp * ΔT
Where:
Q2 is the heat lost by the water (in Joules)
m2 is the mass of the water (in grams)
Cp is the specific heat capacity of water, which is 4.18 J/g·°C
ΔT is the change in temperature (in °C)
We need to find the mass of the water, so let's determine ΔT first. ΔT is the difference between the initial and final temperatures:
ΔT = 9.0°C - 0.0°C
ΔT = 9.0°C
Now we substitute the values into the formula:
Q2 = m2 * 4.18 J/g·°C * 9.0°C
Q2 = 37.62 m2
Since the total heat lost by the water (Q2) is equal to the heat gained by the ice (Q1) according to the principle of conservation of energy, we can set up the equation:
Q1 = Q2
16366 J = 37.62 m2
To solve for m2, we divide both sides of the equation by 37.62:
m2 = 16366 J / 37.62
m2 ≈ 435.6 g
Therefore, the mass of water in the sample is approximately 435.6 grams.
To solve this problem, we need to use the principle of conservation of energy. The heat gained by the ice in melting is equal to the heat lost by the water in cooling.
First, let's calculate the heat gained by the ice in melting. The heat added or lost can be calculated using the equation:
Q = m * L
Where:
Q is the heat added or lost
m is the mass of the substance (ice or water)
L is the latent heat of fusion for water, which is 334 J/g
In this case, the ice is at 0.0°C and is completely melted, so the heat gained by the ice is:
Q_ice = m_ice * L
Next, let's calculate the heat lost by the water in cooling. The heat gained or lost can be calculated using the equation:
Q = m * c * (ΔT)
Where:
Q is the heat added or lost
m is the mass of the substance (ice or water)
c is the specific heat capacity of water, which is 4.18 J/g°C
ΔT is the change in temperature
In this case, the water temperature decreases from 9.0°C to 0.0°C, so the heat lost by the water is:
Q_water = m_water * c * (ΔT)
Since the heat gained by the ice is equal to the heat lost by the water, we can set up an equation:
Q_ice = Q_water
m_ice * L = m_water * c * (ΔT)
Now we can solve for the mass of water, m_water, using the values given in the problem:
49g * 334 J/g = m_water * 4.18 J/g°C * (9.0°C - 0.0°C)
Simplifying the equation:
16366 J = m_water * 4.18 J/g°C * 9.0°C
Dividing both sides of the equation by (4.18 J/g°C * 9.0°C):
m_water = 16366 J / (4.18 J/g°C * 9.0°C)
m_water ≈ 463.38 g
Therefore, there were approximately 463.38 grams of water in the sample.
(mass ice x heat fusion) + [mass water x specific heat water x (Tfinal-Tinitial)] = 0
Solve for mass water.