A 62g piece of ice at 0 degrees celcius is added to a sample of water at 9 degrees celcius. All of the ice melts and the temperature of the water decreases to 0 degrees celcius. How many grams of water were in the sample
(mass ice x heat fusion) + (mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0
Solve for mass H2O.
Note the correct spelling of celsius.
To solve this problem, we can use the principle of conservation of energy. The heat gained by the water is equal to the heat lost by the ice during this process. We can calculate the heat gained/lost using the formula:
Q = mcΔT
Where:
Q is the heat gained/lost (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
For the water, the heat gained is:
Q_water = m_water * c_water * ΔT_water
For the ice, the heat lost is:
Q_ice = m_ice * c_ice * ΔT_ice
Since the final temperature is 0°C, the heat gained by the water is equal to the heat removed from the ice:
Q_water = Q_ice
Now let's calculate the heat gained by the water:
Q_water = m_water * c_water * ΔT_water
Q_water = m_water * 4.18 J/g°C * (0°C - 9°C)
Q_water = m_water * 4.18 J/g°C * -9°C
Q_water = -37.62 m_water J
Since Q_water = Q_ice, we can equate the two equations:
-37.62 m_water = m_ice * 2.09 J/g°C * (0°C - 0°C)
-37.62 m_water = 0
Since m_water cannot be negative, we know that m_ice must be 0. Therefore, there was no ice present, and the entire sample was water.
Therefore, the mass of water in the sample is 62 grams.