a 40g piece of ice at 0.0 degrees C is added to a sample of water at 8.0 degrees C.
the ice melts and the temp of the water decreses to 0.0 degrees C.
how many kilograms of water weere in the sample?
[mass ice x heat fusion] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for mass H2O.
To solve this problem, we can use the concept of heat transfer. The amount of heat gained by the ice is equal to the amount of heat lost by the water.
1. Calculate the heat gained by the ice:
The specific heat capacity of ice is 2.09 J/g°C, and the temperature change is 0.0°C to 0.0°C (no change in temperature since the ice is melting). Therefore, the heat gained by the ice is 0 J.
2. Calculate the heat lost by the water:
The specific heat capacity of water is 4.18 J/g°C, and the temperature change is 8.0°C to 0.0°C. Therefore, the amount of heat lost by the water can be calculated using the formula:
Q = m * c * ΔT
Where:
Q = heat lost
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
Substitute the values into the formula:
Q = m * 4.18 * (8.0 - 0.0)
Q = 33.44 m
3. Since the heat gained by the ice is equal to the heat lost by the water, we can equate them:
0 = 33.44 m
Solve for the mass of water (m):
m = 0 / 33.44
m = 0 kg
Therefore, there was 0 kilograms of water in the sample.
Note: The result of this calculation suggests that there may be a mistake or a missing piece of information in the problem statement, as it is not possible to have a decrease in temperature of the water without any heat exchange.