a 42 piece of ice at 0.0 C is added to a sample of water at 8.0 C. all of the ice melts and the temperature of the water decreases to 0.0 C. how many grams of water were in the sample.
To find the number of grams of water in the sample, we need to know the heat transfer that occurred during the process. We can use the equation:
q = m * C * ΔT
where:
q is the heat transferred
m is the mass of the substance
C is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's calculate the heat transferred when the ice melted:
q1 = m1 * C1 * ΔT1
Since the ice melted at 0.0°C, ΔT1 = 0.0 - (-10.0) = 10.0°C (assuming the initial temperature of the ice was -10.0°C)
The specific heat capacity of ice (C1) is approximately 2.09 J/g°C.
Given that q1 = 42 pieces * 2.09 J/g°C * 10.0°C = 878.4 J
Now, let's calculate the heat transferred when the melted ice cooled the water:
q2 = m2 * C2 * ΔT2
Since the water cooled from 8.0°C to 0.0°C, ΔT2 = 0.0 - 8.0 = -8.0°C
The specific heat capacity of water (C2) is approximately 4.18 J/g°C.
Given that q2 = m2 * 4.18 J/g°C * (-8.0)°C = -33.44 m2
Since the heat lost by the water is equal to the heat gained by the ice:
q1 + q2 = 0
878.4 J - 33.44 m2 = 0
Solving for m2:
m2 = 878.4 J / 33.44 J/g = 26.2 g
Therefore, there were approximately 26.2 grams of water in the sample.
melt ice + move ice from zero to final T + cool H2O from 8 to final T = 0
(mass ice x heat fusion) + (mass melted ice x specific heat H2O x (Tfinal-Tinitial) + (mass H2O x speicif heat H2O x (Tfinal-Tinitial) = 0
Solve for mass H2O.
I obtained 420 g but you should confirm that value.